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Question: The equilibrium constant for the reaction, SO3(g) \(\rightleftharpoons\) SO2(g) + \(\frac{1}{2}\)O...

The equilibrium constant for the reaction, SO3(g) ⇌\rightleftharpoons SO2(g) + 12\frac{1}{2}O2(g)

A

416

B

2.40 × 10–3

C

9.8 × 10–2

D

4.9 × 10–2

Answer

416

Explanation

Solution

SO3(g) ⇌\rightleftharpoons SO2(g) + O2(g)

[SO2][O2]1/2[SO3]=KC=4.9×10−2\frac{\left\lbrack SO_{2} \right\rbrack\left\lbrack O_{2} \right\rbrack^{1/2}}{\left\lbrack SO_{3} \right\rbrack} = K_{C} = 4.9 \times 10^{- 2} ..........(i)

SO3(g)+1/25O2SO_{3}(g) + 1/25O_{2} ⇌\rightleftharpoons SO3(g)\text{S}\text{O}_{3}(g) .........(ii)[SO3][SO2][O2]1/2=K′C=14.9×10−2\frac{\left\lbrack SO_{3} \right\rbrack}{\left\lbrack SO_{2} \right\rbrack\left\lbrack O_{2} \right\rbrack^{1/2}} = K'C = \frac{1}{4.9 \times 10^{- 2}}

For 2SO2(g) + O2(g) 2SO3(g)

14.9×10−2=KC2=14.9×4.9×10−4=1000024.01=416.49\frac{1}{4.9 \times 10^{- 2}} = K_{C}2 = \frac{1}{4.9 \times 4.9 \times 10^{- 4}} = \frac{10000}{24.01} = 416.49