If the reaction SO2(g) +0.5O2(g) <-> SO3(g) has the equilibrium constant Kc =56, then what is the Kc value for the following reaction?
2SO3 (g) <-> 2SO2 (g) + O2(g)
A. -112
B. 3.2 * 10^-4
C. 56
D. 8.9*10^-3
I think the answer would be C because the amount of reactants and products are equal and so the equilibrium constant should be the same. Can someone verify the answer please.
2 answers
Well, I don't know chemistry, but, do you feel confident about your answer?
No, C is not correct for two reasons.
First, the reaction in the question is the reverse of that shown.
Second, the coefficients are different
Therefore you must make two corrections to the Kc = 56.
To correct for the fact that it is reversed the new Kc will be 1/Kc or 1/56.
Then to that you correct for the fact that the coefficients are twice the original you square that so the final answer is (1/56)^2. I'm guessing that will be B/
And to Double Pi if you're reading this, I should like to suggest that you cool it. If you don't answer a question we will know you don't know the answer. And for many of the ones you have answered you have done so incorrectly. So, to keep from cluttering the board with useless information, please refrain from answering unless you are positive you know the answer. Thank you.
First, the reaction in the question is the reverse of that shown.
Second, the coefficients are different
Therefore you must make two corrections to the Kc = 56.
To correct for the fact that it is reversed the new Kc will be 1/Kc or 1/56.
Then to that you correct for the fact that the coefficients are twice the original you square that so the final answer is (1/56)^2. I'm guessing that will be B/
And to Double Pi if you're reading this, I should like to suggest that you cool it. If you don't answer a question we will know you don't know the answer. And for many of the ones you have answered you have done so incorrectly. So, to keep from cluttering the board with useless information, please refrain from answering unless you are positive you know the answer. Thank you.