I was wondering about adding for delta H, and delta S. Sometimes I mix the numbers up and put positives where there should be negatives. for example number 11 on summer 2007 test. for the delta H I got all the numbers right except for PH3(g). How come it is positive not negative?
Calculate the following values for the unbalanced reaction listed at 25ºC.
Reaction 1: P4O10(s) + H2(g) -> PH3(g) + H2O(g)
The values listed in the table are (in order of the reaction):
{delta}Hf : -2984 ; 0 ; +5.4 ; -241.818
So : +228.86 ; +130.684 ; +310.23 ; +188.825
{delta}Gf : -2697.7 ; 0 ; +13.4 ; -228.572
If something is a reactant, change the sign listed in the table; if it's a product, don't change it. PH3(g) is a product in this reaction, so use the values directly from the table, in this case they all happen to be positive numbers. You can think of these numbers sort of like prices. Let's say you look up the price of a calculator and it's $10. If you're buying the calculator, the balance in your checking account will decrease by $10. If you're selling the calculator, the balance in your checking account will increase by $10. If you're forming something in a reaction, the values listed in the table represent what you have; if you're consuming something in a reaction, the values in the table represent what you have lost.
Someone also asked about a Mastering Chemistry question from Assignment 5. It has 2 parts...
The rate constant for a certain reaction is k = 3.30×10−3 s-1. If the initial reactant concentration was 0.250M, what will the concentration be after 13.0 minutes?
Integrated rate laws are all about choosing the correct order and then doing some algebra. In this case, the order of the rate law expression must be determined by the units of the rate law constant. Remember that:
Rate0 = k[reactant]x
Since the rate is expressed as "M/time", then this must be a first-order rate law because (1/s)(M) = M/s. That means we have to use the first-order integrated rate law expression
ln[reactant]t = -kt + ln[reactant]0
ln[reactant]13.0min = -(3.30x10-3 s-1){(13.0min)(60s/min)} + ln(0.250) = -3.96
[reactant]13.0min = 0.0191M
A zero-order reaction has a constant rate of 4.30×10−4 M/sec. If after 40.0seconds the concentration has dropped to 5.00×10−2 M, what was the initial concentration?
This one tells you it's a zero-order process, but it might trip you up by giving you the "constant rate" rather than a "rate constant". We can plug the rate and the concentration at 40.0sec in to the regular rate law expression to get the rate law constant:
Rate0 = k[reactant]0
(4.30x10-4 M/sec) = k (5.00x10-2 M)0
Since this is a zero-order process, it turns out that k is equal to the constant rate. Again, plug values into the correct integrated rate law expression and have some algebra fun. For a zero-order process, the IRL is:
[reactant]t = -kt + [reactant]0
(5.00x10-2 M) = -(4.30x10-4 M/sec)(40.0sec) + [reactant]0
Solving, the initial concentration of reactant was 0.0672M.
Other questions, let me know...
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