The equilibrium constant of the reaction,
${S O}_{3} (g) ⇌ {S O}_{2} (g) + \frac{1}{2} O_{2} (g)$
is 0. 15 at 900 K. Calculate the equilibrium constant for
$2 {S O}_{2} (g) + O_{2} (g) ⇌ {2 S O}_{3} (g)$ :

Question

The equilibrium constant of the reaction,

{S O}_{3} (g) ⇌ {S O}_{2} (g) + \frac{1}{2} O_{2} (g)

is 0. 15 at 900 K. Calculate the equilibrium constant for

2 {S O}_{2} (g) + O_{2} (g) ⇌ {2 S O}_{3} (g)

:

Toppr Admin · Accepted Answer

SO2-g-1-2O2-g-x21CC-SO3-g-xA0- -xA0- -xA0-K1-4-xD7-10-x2212-3 atm-x2212-0-5-2SO3-g-x21CC-2SO2-g-O2-g-xA0-To get the equilibrium constant of the required equation-K1-SO3-SO2-O2-1-2

The equilibrium constant of the reaction SO2(g)+1/2O2(g)⇌SO3(g) is 4×10−3 atm−0.5. The equilibrium constant of the reaction 2SO3(g)⇌2SO2(g)+O2(g) would be:4×103 atm250 atm6.25×104 atm0.25×104 atm

SO2(g)+1/2O2(g)⇌SO3(g) K1=4×10−3 atm−0.5.2SO3(g)⇌2SO2(g)+O2(g) To get the equilibrium constant of the required equation:K1=[SO3][SO2][O2]1/2

$S O_{2} (g) + 1 / 2 O_{2} (g) ⇌ S O_{3} (g)$ $K_{1} = 4 \times 10^{- 3}$ $a t m^{- 0.5}$ .

$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$

To get the equilibrium constant of the required equation:

$K_{1} = \frac{[S O_{3}]}{[S O_{2}] [O_{2}]^{1 / 2}}$