Given that the equilibrium constant for the reaction 2SO2(g) + O2(g) leftharpoons 2SO3(g) has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature? SO3(g) leftharpoons SO2(g) + (1/2) O2(g)

Question

Given that the equilibrium constant for the reaction 2SO2(g) + O2(g) leftharpoons 2SO3(g) has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature? SO3(g) leftharpoons SO2(g) + (1/2) O2(g) (A) 1.8 × 10-3 (B) 3.6 × 10-3 (C) 6.0 × 10-2 (D) 1.3 × 10-5

Tardigrade Admin · Accepted Answer

2SO2(g) + O2(g)leftharpoons 2SO3(g) K = 278 (given) SO3(g) leftharpoons SO2(g) + (1/2) O2(g) K' = (√(1/K)) = √(1/278) = √35.97×10-4 = 6×10-2

$1.8 \times 1 0^{- 3}$

$3.6 \times 1 0^{- 3}$

$6.0 \times 1 0^{- 2}$

$1.3 \times 1 0^{- 5}$

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g) K = 278$ (given)
$S O_{3} (g) ⇌ S O_{2} (g) + \frac{1}{2} O_{2} (g) K^{'} = (\frac{1}{K})$
$= \frac{1}{278} = 35.97 \times 1 0^{- 4} = 6 \times 1 0^{- 2}$

Q. Given that the equilibrium constant for the reaction 2SO2​(g)+O2​(g)⇌2SO3​(g) has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature? SO3​(g)⇌SO2​(g)+21​O2​(g)

1.8×10−3

3.6×10−3

6.0×10−2

1.3×10−5

2SO2​(g)+O2​(g)⇌2SO3​(g)K=278 (given) SO3​(g)⇌SO2​(g)+21​O2​(g)K′=(K1​​) =2781​​=35.97×10−4​=6×10−2

$1.8 \times 1 0^{- 3}$

$3.6 \times 1 0^{- 3}$

$6.0 \times 1 0^{- 2}$

$1.3 \times 1 0^{- 5}$

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g) K = 278$ (given)
$S O_{3} (g) ⇌ S O_{2} (g) + \frac{1}{2} O_{2} (g) K^{'} = (\frac{1}{K})$
$= \frac{1}{278} = 35.97 \times 1 0^{- 4} = 6 \times 1 0^{- 2}$