Asked by Erka
For the following reaction, 2SO3(g) = 2SO2(g) + O2(g), the equilibrium constant, Kp, is 1.32 at 627 degrees Celsius. What is the equilibrium constant for the reaction: SO3(g) = SO2(g)+ 1/2 O2(g)
Answers
Answered by
bobpursley
One thing that you should note quickly is that the equilibrium constant expression depends on how we write the chemical reaction.
For example, consider the following reaction
N2O5(g) < - > 2NO2(g) + 1/2O2(g) K = [NO2]2[O2]1/2/[N2O5]
However, we can just as easily multiply the whole reaction by 2- perhaps we don't like the stoichiometric coefficient of 1/2 in front of the oxygen. This gives
2N2O5(g) < - > 4NO2(g) + O2(g) K' = [NO2]4[O2]/[N2O5]2
K depends on how you write the chemical reaction! If you multiply the chemical equation by a given number (such as 2), it is the same as raising the equilibrium constant expression to that power. (K' = K2 in the above example.)
For example, consider the following reaction
N2O5(g) < - > 2NO2(g) + 1/2O2(g) K = [NO2]2[O2]1/2/[N2O5]
However, we can just as easily multiply the whole reaction by 2- perhaps we don't like the stoichiometric coefficient of 1/2 in front of the oxygen. This gives
2N2O5(g) < - > 4NO2(g) + O2(g) K' = [NO2]4[O2]/[N2O5]2
K depends on how you write the chemical reaction! If you multiply the chemical equation by a given number (such as 2), it is the same as raising the equilibrium constant expression to that power. (K' = K2 in the above example.)
Answered by
Cin
.870
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