Question

2s(s) + 2o₂(g) ⇌ 2so₂(g) k₁ = 2×10¹⁰⁵
2so₂(g) + o₂(g) ⇌ 2so₃(g) k₂ = 7×10²⁴
given the value of the equilibrium constants k₁ and k₂ for the reactions represented above, what is the value of the equilibrium constant, k₃, for the following reaction?
2s(s) + 3o₂(g) ⇌ 2so₃(g)
a. 1×10¹³⁰
b. 3×10⁶⁰
c. 1×10⁶⁵
d. 2×10⁴⁰
e. 7×10²⁴

Explanation:

Step1: Combine the two given reactions

The first reaction is \(2S(s)+2O_{2}(g)
ightleftharpoons 2SO_{2}(g)\) with \(K_{1} = 2\times10^{105}\). The second reaction is \(2SO_{2}(g)+O_{2}(g)
ightleftharpoons 2SO_{3}(g)\) with \(K_{2}=7\times 10^{24}\). When we add these two reactions together (\(2S(s)+2O_{2}(g)+2SO_{2}(g)+O_{2}(g)
ightleftharpoons 2SO_{2}(g)+2SO_{3}(g)\)), the \(2SO_{2}(g)\) cancels out on both sides, and we get \(2S(s)+3O_{2}(g)
ightleftharpoons 2SO_{3}(g)\) which is the target reaction.

Step2: Use the rule for equilibrium constants of combined reactions

For two reactions \(A
ightleftharpoons B\) with \(K_{a}\) and \(B
ightleftharpoons C\) with \(K_{b}\), when we combine them \(A
ightleftharpoons C\), the equilibrium constant \(K = K_{a}\times K_{b}\). So for our case, \(K_{3}=K_{1}\times K_{2}\).
Substitute \(K_{1} = 2\times10^{105}\) and \(K_{2}=7\times 10^{24}\) into the formula: \(K_{3}=(2\times10^{105})\times(7\times 10^{24})\).
Using the rule of exponents \(a^{m}\times a^{n}=a^{m + n}\) and \(c\times10^{m}\times d\times10^{n}=(c\times d)\times10^{m + n}\), we have \(K_{3}=(2\times7)\times10^{105 + 24}=14\times10^{129}\approx1\times 10^{130}\) (since \(14\approx10\) when considering the order - of - magnitude approximation for equilibrium constant values in this context).

Answer:

A. \(1\times 10^{130}\)