Calculate the equilibrium constant for the reaction H_2(g) + CO_2(g) H_2O(g) + CO(g) at 1395 ,K by using following datas 2H_2O(g) 2H_2(g) + O_2(g); K_1 = 2.1 10^ -13 2CO_2(g) 2CO(g) + O_2(g); K_2 = 1.4 10^ -12

Calculate the equilibrium constant for the reaction H_2(g) + CO_2…

MCQ

Calculate the equilibrium constant for the reaction $H_2(g) + CO_2(g) \rightleftharpoons H_2O(g) + CO(g)$ at $1395 \,K$ by using following datas

$2H_2O(g) \rightleftharpoons 2H_2(g) + O_2(g); K_1 = 2.1 \times 10^{-13}$

$2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g); K_2 = 1.4 \times 10^{-12}$

A
$1.27$
✓
$2.58$
C
$4.17$
D
$4.93$

✓

Answer

Correct option: B.

$2.58$

b
For $2 H _2 O _{( g )} \rightleftharpoons 2 H _{2( g )}+ O _{2( g )} ; K _1=\frac{\left[ H _2\right]^2\left[ O _2\right]}{\left[ H _2 O \right]^2} \ldots .(1)$

For $2 CO _{2( g )} \rightleftharpoons 2 CO _{( g )}+ O _{2( g )}; K _2=\frac{[ CO ]^2\left[ O _2\right]}{\left[ CO _2\right]^2} \ldots(2)$

For $CO _{2( g )}+ H _{2( g )} \rightleftharpoons H _2 O _{( g )}+ CO _{( g )}; K =\frac{\left[ H _2 O \right][ CO ]}{\left[ CO _2\right]\left[ H _2\right]} \ldots(3)$

By dividing eq. $(2)$ by eq. $(1):$

$\frac{ K _2}{ K _1}=\frac{[ CO ]^2\left[ O _2\right]}{\left[ CO _2\right]^2} \times \frac{\left[ H _2 O \right]^2}{\left[ H _2\right]^2\left[ O _2\right]}$

$\frac{ K _2}{ K _1}=\frac{[ CO ]^2\left[ H _2 O \right]^2}{\left[ CO _2\right]^2\left[ H _2\right]^2}= K ^2 \text { by Eq. (3) }$

or $K =\sqrt{\left(\frac{ K _2}{ K _1}\right)}=\sqrt{\left(\frac{1.4 \times 10^{-12}}{2.1 \times 10^{-13}}\right)}=2.58$