In exam #3 question #1 how do you find the H+ or OH- from pOH of 6.113?
pH + pOH = pKw = 14 @25degC
So, if pOH = 6.113, pH = 7.887
[H+] = 10-pH = 10-7.887 = 1.30x10-8 M
[H+] [OH-] = Kw = 10-14 @25degC so
(1.30x10-8)[OH-] = 10-14
[OH-] = 7.71x10-7 M
As a check,
pOH = -log[OH-] = -log(7.71x10-7) = 6.113
2009-07-22
Questions...
On exam #4 question #6: For each of the following reactions, predict whether the sign if delta S will be positive or negative and explain your answer. What are you looking for as an answer for the explanation part?
In the first one, I expected to see something like "A solid is forming from two solutions, so the products are more ordered/less disordered than the reactants." For the second, 2 molecules of gas are forming; for the third, the number of gas particles is changing. A number of people were giving explanations like "positive because the reaction is getting more disordered." That's not an explanation. If your answer to an "explain" question does not address the question "Why?", then it's probably not really explaining the answer.
In exam #3 question #1 could you explain again how to find H+ and OH- as well as how to find Ka and Kb?
[H+] and [OH-] are related by Kw. Ka and Kb are also related by Kw.
[H+] [OH-] = Kw
(Ka)(Kb) = Kw
At 25degC, Kw = 10-14
In exam #3 question #2: 1.28 mol K2SO3 + 1.24 mol HCl, can you explain why this does not result in an effective buffer?
This will not make an effective buffer because adding 1.24mols of HCl will almost bring you to the equivalence point in this titration, it will not give you a buffer. You could make a buffer by adding 0.64mol of HCl(aq) to 1.28mols of K2SO3(aq) because the resulting solution would contain 0.64mols of SO32-(aq) {a weak base} and 0.64mols of HSO3-(aq) {its weak conjugate acid}. You could also make an effective buffer by adding 1.92mols of HCl to 1.28mols of K2SO3(aq) because the resulting solution would contain 0.64mols of HSO3-(aq) {a weak base} and 0.64mols of H2SO3(aq) {its weak conjugate acid}.
In exam #2 could you run through #12: Graphite (solid carbon) reacts with oxygen gas to form carbon monoxide gas. You have sealed 11.37g of graphite and 18.61g of oxygen in a 6.00L vessel and allowed the system to reach equilibrium at 73.91C. if the equilibrium constant value is 9.42x10-6 at this temperature, what are the equilibrium concentrations of all reactants and products?
Start with a balanced chemical equation:
C(s) + O2(g) <=> CO2(g)
Since carbon is a solid in this reaction, it does not appear in the equilibrium constant expression, so we can put together a table {again, formatting tables in this program is rough, so I'll simply list the concentrations}:
[O2]initial = 18.61g / 31.998g/mol / 6.00L = 0.0969M
[CO2]initial = 0M
[O2]change = -xM
[CO2]change = +xM
[O2]equilibrium = (0.0969-x)M
[CO2]equilibrium = xM
Kc = [CO2]equilibrium / [O2]equilibrium = (x) / (0.0969-x) = 9.42x10-6
This isn't a horrible equation to solve directly, but let's see if we can make an assumption. If "x" is much smaller than 0.0969M, then this expression simplifies to:
(x) / 0.0969 = 9.42x10-6
x = 9.13x10-7
{"x" is indeed much smaller than 0.0969, so our assumption is valid.}
[O2]equilibrium = 0.0969 M
[CO2]equilibrium = 9.13x10-7 M
In the first one, I expected to see something like "A solid is forming from two solutions, so the products are more ordered/less disordered than the reactants." For the second, 2 molecules of gas are forming; for the third, the number of gas particles is changing. A number of people were giving explanations like "positive because the reaction is getting more disordered." That's not an explanation. If your answer to an "explain" question does not address the question "Why?", then it's probably not really explaining the answer.
In exam #3 question #1 could you explain again how to find H+ and OH- as well as how to find Ka and Kb?
[H+] and [OH-] are related by Kw. Ka and Kb are also related by Kw.
[H+] [OH-] = Kw
(Ka)(Kb) = Kw
At 25degC, Kw = 10-14
In exam #3 question #2: 1.28 mol K2SO3 + 1.24 mol HCl, can you explain why this does not result in an effective buffer?
This will not make an effective buffer because adding 1.24mols of HCl will almost bring you to the equivalence point in this titration, it will not give you a buffer. You could make a buffer by adding 0.64mol of HCl(aq) to 1.28mols of K2SO3(aq) because the resulting solution would contain 0.64mols of SO32-(aq) {a weak base} and 0.64mols of HSO3-(aq) {its weak conjugate acid}. You could also make an effective buffer by adding 1.92mols of HCl to 1.28mols of K2SO3(aq) because the resulting solution would contain 0.64mols of HSO3-(aq) {a weak base} and 0.64mols of H2SO3(aq) {its weak conjugate acid}.
In exam #2 could you run through #12: Graphite (solid carbon) reacts with oxygen gas to form carbon monoxide gas. You have sealed 11.37g of graphite and 18.61g of oxygen in a 6.00L vessel and allowed the system to reach equilibrium at 73.91C. if the equilibrium constant value is 9.42x10-6 at this temperature, what are the equilibrium concentrations of all reactants and products?
Start with a balanced chemical equation:
C(s) + O2(g) <=> CO2(g)
Since carbon is a solid in this reaction, it does not appear in the equilibrium constant expression, so we can put together a table {again, formatting tables in this program is rough, so I'll simply list the concentrations}:
[O2]initial = 18.61g / 31.998g/mol / 6.00L = 0.0969M
[CO2]initial = 0M
[O2]change = -xM
[CO2]change = +xM
[O2]equilibrium = (0.0969-x)M
[CO2]equilibrium = xM
Kc = [CO2]equilibrium / [O2]equilibrium = (x) / (0.0969-x) = 9.42x10-6
This isn't a horrible equation to solve directly, but let's see if we can make an assumption. If "x" is much smaller than 0.0969M, then this expression simplifies to:
(x) / 0.0969 = 9.42x10-6
x = 9.13x10-7
{"x" is indeed much smaller than 0.0969, so our assumption is valid.}
[O2]equilibrium = 0.0969 M
[CO2]equilibrium = 9.13x10-7 M
2009-07-19
Sorry, I had a network problem...
Could you go over #2 & #3 from summer 2009 problem set #11?
Thanks
#2 is an algebra problem.
NO2(g) + 2 H2S(g) <=> NS2(g) + 2 H2O(g)
ΔGrxn = (1)(-ΔGf(NO2(g))) + (2)(-ΔGf(H2S(g))) + (1)(ΔGf(NS2(g))) + (2)(ΔGf(H2O(g)))
From the problem and the tables in your book, you know all of the values in this equation except ΔGf(NS2(g)). Plug in and solve for ΔGf(NS2(g)).
#3 has a couple parts. First, write out a balanced net ionic equation, it will be difficult to handle this problem if you use a full molecular equation. This is just a backwards Ksp process, so the net ionic equation is:
Pb2+(aq) + 2 Cl-(aq) <=> PbCl2(s)
Calculate ΔG for this reaction at standard conditions from tabulated values. {I don't have a book with tabulated thermodynamic values handy so I can't calculate an exact value for you...} Once you know ΔGo, plug in to calculate under non-standard conditions.
ΔG = ΔGo + RTlnQ
Q = 1 / {[Pb2+][Cl-]2}
Given the numbers in the problem, the correction term {RTlnQ} will probably only be a few kJ/mol. Remember to use kelvin temperatures and be consistent with kJ and J.
Thanks
#2 is an algebra problem.
NO2(g) + 2 H2S(g) <=> NS2(g) + 2 H2O(g)
ΔGrxn = (1)(-ΔGf(NO2(g))) + (2)(-ΔGf(H2S(g))) + (1)(ΔGf(NS2(g))) + (2)(ΔGf(H2O(g)))
From the problem and the tables in your book, you know all of the values in this equation except ΔGf(NS2(g)). Plug in and solve for ΔGf(NS2(g)).
#3 has a couple parts. First, write out a balanced net ionic equation, it will be difficult to handle this problem if you use a full molecular equation. This is just a backwards Ksp process, so the net ionic equation is:
Pb2+(aq) + 2 Cl-(aq) <=> PbCl2(s)
Calculate ΔG for this reaction at standard conditions from tabulated values. {I don't have a book with tabulated thermodynamic values handy so I can't calculate an exact value for you...} Once you know ΔGo, plug in to calculate under non-standard conditions.
ΔG = ΔGo + RTlnQ
Q = 1 / {[Pb2+][Cl-]2}
Given the numbers in the problem, the correction term {RTlnQ} will probably only be a few kJ/mol. Remember to use kelvin temperatures and be consistent with kJ and J.
Could you go over #2 & #3 from summer 2009 problem set #11?
Thanks
Thanks
#2 is an algebra problem.
NO2(g) + 2 H2S(g) <=> NS2(g) + 2 H2O(g)
ΔGrxn = (1)(-ΔGf(NO2(g))) + (2)(-ΔGf(H2S(g))) + (1)(ΔGf(NS2(g))) + (2)(ΔGf(H2O(g)))
From the problem and the tables in your book, you know all of the values in this equation except ΔGf(NS2(g)). Plug in and solve for ΔGf(NS2(g)).
#3 has a couple parts. First, write out a balanced net ionic equation, it will be difficult to handle this problem if you use a full molecular equation. This is just a backwards Ksp process, so the net ionic equation is:
Pb2+(aq) + 2 Cl-(aq) <=> PbCl2(s)
Calculate ΔG for this reaction at standard conditions.
ΔG = ΔGo + RTlnQ
ΔG = ΔGo + RTlnQ
Q = 1 / {[Pb2+][Cl-]2}
Questions...
Can you explain again why certain values are negative or positive like in the spring/winter exam reaction #2
Fe(CO)6(l) + ClO2(g) FeCl3(s) + CO2(g)
ΔHrxnº = (783.5 kJ/mol) + 3(-102.5 kJ/mol) + (-399.49 kJ/mol) + 6(-393.509 kJ/mol) = -2284.5 kJ/mol
The ΔH and ΔG values you find in the tables are formation values. If something is being formed in a reaction, the values are correct as you find them in the table. If something in being consumed/desstroyed/used up in a reaction, the magnitude of the value in the table is correct, but the sign is wrong. If something is a reactant, change the sign; if it's a product, the sign in the table is correct.
I was wondering if you could answer #3 from problem set 10?
What is the molecular basis of enthalpy, entropy, and free energy? Let's start with entropy. Gaseous water is more disordered than liquid water, so we expect the entropy to be higher. For both enthalpy and free energy, gaseous water is more energetic than liquid water, so whether we're talking about heat (enthalpy) or overall energy (free energy), we expect the gas to be higher energy than liquid. Note, in this case, "higher" energy means "less negative" since all of these numbers are negative.
Other questions, let me know...
Fe(CO)6(l) + ClO2(g) FeCl3(s) + CO2(g)
ΔHrxnº = (783.5 kJ/mol) + 3(-102.5 kJ/mol) + (-399.49 kJ/mol) + 6(-393.509 kJ/mol) = -2284.5 kJ/mol
The ΔH and ΔG values you find in the tables are formation values. If something is being formed in a reaction, the values are correct as you find them in the table. If something in being consumed/desstroyed/used up in a reaction, the magnitude of the value in the table is correct, but the sign is wrong. If something is a reactant, change the sign; if it's a product, the sign in the table is correct.
I was wondering if you could answer #3 from problem set 10?
What is the molecular basis of enthalpy, entropy, and free energy? Let's start with entropy. Gaseous water is more disordered than liquid water, so we expect the entropy to be higher. For both enthalpy and free energy, gaseous water is more energetic than liquid water, so whether we're talking about heat (enthalpy) or overall energy (free energy), we expect the gas to be higher energy than liquid. Note, in this case, "higher" energy means "less negative" since all of these numbers are negative.
Other questions, let me know...
2009-07-15
PS#11, July 15th...
Chem 210 – Summer 2009 – Problem Set #11
1. You have combined 50.0mL of 0.927M barium nitrate solution and 50.0mL of 0.899M sodium sulfate solution to form barium sulfate. You have captured all of the energy liberated by this spontaneous process to decompose water to hydrogen gas and oxygen gas. How many grams of hydrogen gas could you produce?
2. You are attempting to find the Gibb’s free energy of formation for NS2(g). When you react nitrogen dioxide with H2S(g) to produce NS2(g) and gaseous water, the free energy change for the reaction is 28.7kJ/mol. What is ΔG°f for NS2(g)?
3. You have reacted lead(II) nitrate solution and sodium chloride solution under standard conditions to produce lead(II) chloride. What is ΔG° for this reaction? What is ΔG for this reaction if you use 0.85M lead(II) nitrate solution, 1.7M sodium chloride solution, and perform the reaction at 15.00°C? At 91.82°C?
PS#10 from Tuesday, July 14th
Chem 210 – Summer 2009 – Problem Set #10
1. Ethanol {C2H5OH(l)} reacts with oxygen gas to create carbon dioxide and water.
a. Calculate ΔH°, ΔS°, and ΔG° for this reaction. Is this reaction spontaneous?
b. How much energy is transferred between the system and the surroundings when 10.83g of ethanol is burned in excess oxygen? Is energy transferred from the system to the surroundings, or from the surroundings to the system?
2. Ideally, the hydrogen gas used in fuel cells would come from water.
a. Calculate ΔH°, ΔS°, and ΔG° for the decomposition of water to hydrogen gas and oxygen gas. Is this reaction spontaneous?
b. How much energy is transferred between the system and the surroundings when 6.28g of hydrogen gas is produced by this reaction? Is energy transferred from the system to the surroundings, or from the surroundings to the system?
3. In the previous problems, did you use liquid water or gaseous water? Why? How will changing from liquid-to-gas or gas-to-liquid change your answers? On a molecular level, explain the differences in ΔH°f, S°, and ΔG°f for liquid and gaseous water using the numbers found in thermodynamics tables. Why is gaseous water higher or lower than liquid water for each thermodynamic quantity?
4. If burning ethanol is a step in a reaction mechanism, the reaction may have to be multiplied by some integer.
a. Double all the coefficients in the balanced equation from #1 and calculate ΔG° for the “new” reaction. How is this value of ΔG° related to the value calculated in #1?
b. How much energy is transferred between the system and the surroundings when 10.83g of ethanol is burned in excess oxygen? Compare your answer to the value you found in #1b.
5. If 50.00g of ethanol is burned and all of the water produced by the reaction is decomposed to form hydrogen gas, what is the change in Gibb’s free energy for the whole process?
2009-07-12
Question...
exam #3 from winter 2006
Question #2 - could you explain how to determine if it is an effective buffer for each combination listed
Thank you
2. Does the combination listed result in an effective buffer solution? (4pts each)
Yes No 0.38mol HCl(aq) + 0.38mol NaOH(aq)
Yes No 0.90mol Na3PO4(aq) + 1.35mol HNO3(aq)
Yes No 1.28mol Na2CO3(aq) + 0.64mol HCl(aq)
Yes No 2.14mol CH3COOH(aq) + 1.96mol CH3COOK(aq)
Yes No 0.06mol HCN(aq) + 0.98mol LiCN(aq)
An effective buffer contains approximately equal amounts of a weak acid and its conjugate base with the concentration of each at least 100x the Ka of the conjugate acid. So let's look at each combination:
0.38mol HCl(aq) + 0.38mol NaOH(aq)
This is a strong acid and a strong base, the result of this combination will give a solution of sodium chloride in water. Not a buffer.
0.90mol Na3PO4(aq) + 1.35mol HNO3(aq)
Nitric acid is a strong acid, so when 1.35mol of strong acid are added to the phosphate (a weak base), the result of this combination will be 0.45mols of HPO42-(aq) and H2PO4-(aq). This is a combination of a weak conjugate acid and its weak conjugate base. This will be a good buffer.
1.28mol Na2CO3(aq) + 0.64mol HCl(aq)
Similar to the previous example, the result of mixing these solutions together will be a combination of 0.64mols HCO3-(aq) and CO32-(aq). This will be a good buffer.
2.14mol CH3COOH(aq) + 1.96mol CH3COOK(aq)
Acetate ion is the conjugate base of acetic acid (weak acid), and this is an approximately equimolar mixture of acetic acid and acetate ion. This will be a good buffer.
0.06mol HCN(aq) + 0.98mol LiCN(aq)
HCN is a weak acid and cyanide ion is its conjugate base, but the concentrations are too far apart to make an effective buffer. The concentration of the conjugate acid and conjugate base should be within a factor of 10 to make as effective buffer. This is not an effective buffer
2009-07-11
Email question...
I'm looking at Exam 3 from Winter 2006, and for question 5, shouldn't K2SO3 be the one being added to HClO4 since it is stonger?
Perchloric acid {HClO4(aq)} is probably one of the strongest acids we'll consider, so it's pretty safe to say that any titration involving perchloric acid will have the perchloric acid in the buret.
Again, when you are setting up a titration it is always preferable to have a something strong and monoprotic in the buret.
Other questions, let me know, I'll answer them to the blog as soon as I see them...
2009-07-10
Problem Set #9, question #5...
5. You would like to perform a titration using cyanic acid { HCNO(aq), Ka = 2x10-4 } and potassium hydroxide. Which solution should be in the buret? You do not have a pH meter, but you have found a number of indicators. Which of the following would be most appropriate for this titration and why? Indicator: Endpoint range - 2,4-dinitrophenol: 2.8-4.0 ; phenol red: 6.7-8.1 ; thymolphthalein: 9.6-10.3
Although this question has some numbers in it, you don't actually need to do any calculations to answer it. Based upon the Ka value, HCNO(aq) is a weak acid. When performing a titration, you always want to put something strong and monoprotic in the buret. HCNO(aq) is monoprotic, but it is weak. Potassium hydroxide is monoprotic and strong, so it would be the better choice to put in the buret. Now, what about the indicator? When titrating a weak monoprotic acid with a strong monoprotic base, the equivalence point should be somewhere on the basic side, so we should choose an indicator that changes color at basic pH. The range for phenol red includes some slightly basic pH's so it should work, but it's pretty close to neutral, so thymolphthalein is probably the best bet for this titration.
We can also look at this by doing a calculation. When we get to the equivalence point, we will (in essence) have a solution of potassium cyanate in water. If we assume that the concentration of cyanate is 1M, we can calcuate an expected pH of the solution using a Kb-type of equation. {I just pulled 1M out of my hat, you can do this calculation with any concentration to check...}
CNO-(aq) + H2O(l) <=> HCNO(aq) + OH-(aq)
[CNO-]initial = 1 M
[HCNO]initial = 0 M
[OH-]initial = 10-7 M
[CNO-]change = -x M
[HCNO]change = +x M
[OH-]change = +x M
[CNO-]eq = (1-x) M
[HCNO]eq = x M
[OH-]eq = (10-7+x)M
Let's assume that x is significantly smaller than 1 and significantly larger than 10-7, then our Kb expression is:
Kb = {[HCNO]eq [OH-]eq} / [CNO-]eq = { (x)(x) } / 1 = x2 = 5x10-11
NOTE: Kb for CNO- is calculated from the given Ka of HCNO
x = 7.07x10-6
Hmm, this is a pretty small number, our assumption that x is significantly larger than 10-7 might be a little questionable. 10-7 is only about 1.4% of 7.07x10-6, so we should be OK...
x = 7.07x10-6 = [OH-]eq
pOH = -log [OH-] = -log(7.07x10-6) = 5.15
At 25 degC,
pH = 14 - pOH = 8.849
So it looks like the pH at the equivalence point should be basic enough that thymolphthalein will be a better indicator. For practice, redo this calculation assuming that the concentration of potassium cyanate is 0.5M at the equivalence point...
2009-07-05
From a problem set...
Problem Set #4, Question 2:
2. Carbon dioxide reacts with ammonia (NH3) in the gas phase to produce formamide (HCONH2) and oxygen. If 15.215g of CO2(g) and 4.139g of ammonia are combined in a 3.50L vessel and allowed to react equilibrium, the formamide concentration is found to be 24.8mM. What is the equilibrium constant for this reaction? Is this reaction product-favored or reactant-favored?
Balanced equation:
2 CO2(g) + 2 NH3(g) <--> 2 HCONH2(g) + O2(g)
[CO2]initial = 15.215g / 44.009g/mol / 3.50L = 0.098778M
[NH3]initial = 4.139g / 17.031g/mol / 3.50L = 0.069436M
[HCONH2]initial = 0M
[O2]initial = 0M
{NOTE: Once again, I'm carrying too many sig figs here on purpose because this is the middle of the calculation. I'll round later.}
Set these up in a table, I don't have tables active in the blog so....
Change in:
[CO2] = -2x [NH3] = -2x [HCONH2] = +2x [O2] = +x
So at equilibrium,
[CO2]eq = (0.098778-2x)M
[NH3]eq = (0.069436-2x)M
[HCONH2]eq = (0+2x)M
[O2]eq = (0+x)M
The problem states that the equilibrium concentration of formamide is 24.8mM, so we can use that number to find the value of "x".
[HCONH2]eq = (0+2x)M = 24.8mM
x = 12.4mM = 0.0124M
Plugging in to find all equilibrium concentrations:
[CO2]eq = (0.098778-2x)M = (0.098778-0.0248)M = 0.07398M
[NH3]eq = (0.069436-2x)M = (0.069436-0.0248)M = 0.04464M
[HCONH2]eq = (0+2x)M = 0.0248M
[O2]eq = (0+x)M = 0.0124M
Kc = { (0.0248)2(0.0124) } / { (0.07398)2(0.04464)2 } = 0.699
Slightly reactant-favored.
2. Carbon dioxide reacts with ammonia (NH3) in the gas phase to produce formamide (HCONH2) and oxygen. If 15.215g of CO2(g) and 4.139g of ammonia are combined in a 3.50L vessel and allowed to react equilibrium, the formamide concentration is found to be 24.8mM. What is the equilibrium constant for this reaction? Is this reaction product-favored or reactant-favored?
Balanced equation:
2 CO2(g) + 2 NH3(g) <--> 2 HCONH2(g) + O2(g)
[CO2]initial = 15.215g / 44.009g/mol / 3.50L = 0.098778M
[NH3]initial = 4.139g / 17.031g/mol / 3.50L = 0.069436M
[HCONH2]initial = 0M
[O2]initial = 0M
{NOTE: Once again, I'm carrying too many sig figs here on purpose because this is the middle of the calculation. I'll round later.}
Set these up in a table, I don't have tables active in the blog so....
Change in:
[CO2] = -2x [NH3] = -2x [HCONH2] = +2x [O2] = +x
So at equilibrium,
[CO2]eq = (0.098778-2x)M
[NH3]eq = (0.069436-2x)M
[HCONH2]eq = (0+2x)M
[O2]eq = (0+x)M
The problem states that the equilibrium concentration of formamide is 24.8mM, so we can use that number to find the value of "x".
[HCONH2]eq = (0+2x)M = 24.8mM
x = 12.4mM = 0.0124M
Plugging in to find all equilibrium concentrations:
[CO2]eq = (0.098778-2x)M = (0.098778-0.0248)M = 0.07398M
[NH3]eq = (0.069436-2x)M = (0.069436-0.0248)M = 0.04464M
[HCONH2]eq = (0+2x)M = 0.0248M
[O2]eq = (0+x)M = 0.0124M
Kc = { (0.0248)2(0.0124) } / { (0.07398)2(0.04464)2 } = 0.699
Slightly reactant-favored.
Another question...
From Summer 2007, Exam 2:
7. You have found the following value in a table of equilibrium constants:
2 C2H3F3(g) + 3 Cl2(g) 2 C2F3Cl3(g) + 3 H2(g) Kc = 5.19x1018
What is the equilibrium constant for the reaction:
6 C2F3Cl3(g) + 9 H2(g) 6 C2H3F3(g) + 9 Cl2(g)
We're manipulating an equilibrium constant here, let's start with the equilibrium constant expression for the original reaction:
Kc = { [C2F3Cl3]2 [H2]3 } / { [C2H3F3]2 [Cl2]3 } = 5.19x1018
To get the second reaction, we have to reverse the original and multiply it by 3. When we reverse the direction of the equilibrium, the roles of products and reactants change, so the equilibrium constant is inverted. When an equilibrium equation is multiplied by some constant, each of the concentrations is raised to that power, so the whole equilibrium constant expression is raised to that power. This means that the equilibrium constant for the "new" reaction in the problem is:
Kc' = { [C2H3F3]6 [Cl2]9 } / { [C2F3Cl3]6 [H2]9 } = (1 / Kc)3 = 7.15x10-57
Others? Let me know...
7. You have found the following value in a table of equilibrium constants:
2 C2H3F3(g) + 3 Cl2(g) 2 C2F3Cl3(g) + 3 H2(g) Kc = 5.19x1018
What is the equilibrium constant for the reaction:
6 C2F3Cl3(g) + 9 H2(g) 6 C2H3F3(g) + 9 Cl2(g)
We're manipulating an equilibrium constant here, let's start with the equilibrium constant expression for the original reaction:
Kc = { [C2F3Cl3]2 [H2]3 } / { [C2H3F3]2 [Cl2]3 } = 5.19x1018
To get the second reaction, we have to reverse the original and multiply it by 3. When we reverse the direction of the equilibrium, the roles of products and reactants change, so the equilibrium constant is inverted. When an equilibrium equation is multiplied by some constant, each of the concentrations is raised to that power, so the whole equilibrium constant expression is raised to that power. This means that the equilibrium constant for the "new" reaction in the problem is:
Kc' = { [C2H3F3]6 [Cl2]9 } / { [C2F3Cl3]6 [H2]9 } = (1 / Kc)3 = 7.15x10-57
Others? Let me know...
Question...
From Summer 2007, Exam 2:
6. For the reaction:
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
The equilibrium concentrations have been found to be [CO2]eq = 0.568M, [H2O]eq = 0.685M,
[CH4]eq = 1.38x10-8 M, [O2]eq = 0.918 M. What is the equilibrium constant?
I think some of you might be making this harder than it needs to be. The problem is giving you the equilibrium concentrations, so you just need to plug these numbers into the equilibrium constant expression:
Kc = { [CO2] [H2O]2} / { [CH4] [O2]2} = {(0.568)(0.685)2} / {(1.38x10-8)(0.918)2} = 2.29x107
Other questions, let me know...
6. For the reaction:
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
The equilibrium concentrations have been found to be [CO2]eq = 0.568M, [H2O]eq = 0.685M,
[CH4]eq = 1.38x10-8 M, [O2]eq = 0.918 M. What is the equilibrium constant?
I think some of you might be making this harder than it needs to be. The problem is giving you the equilibrium concentrations, so you just need to plug these numbers into the equilibrium constant expression:
Kc = { [CO2] [H2O]2} / { [CH4] [O2]2} = {(0.568)(0.685)2} / {(1.38x10-8)(0.918)2} = 2.29x107
Other questions, let me know...
2009-07-03
Equilibrium Week
This week we discussed equilibrium and a number of specific equilibrium constants including: Kc, Kp, Ksp and Kf. Although these all have specific conditions or reaction types associated with them, they are all equilibrium constants that follow all the same rules and trends.
Equilibrium is on Exam 2 for previous summer and spring exams. In some cases, there might be Ksp questions on Exam 3, but the bulk of equilibrium is on previous Exam 2's.
If you have questions, let me know, I will be checking email over the weekend and I will answer all questions to the blog so everyone can see the answers. Good luck in your studying, and I'll see you Monday.
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