The value of KC for the equilibrium reaction CO2(g) + C(s) 2CO(g) at T( K ) is 0.036 . If the equilibrium concentration of CO 2(g) is 0.004 M, the concentration of CO (g) in mol L -1 is

Question

The value of KC for the equilibrium reaction CO2(g) + C(s) <=> 2CO(g) at T( K ) is 0.036 . If the equilibrium concentration of CO 2(g) is 0.004 M, the concentration of CO (g) in mol L -1 is (A) 3.6 × 10-2 (B) 2.0 × 10-2 (C) 1.2 × 10-2 (D) 1.2 × 10-3

Tardigrade Admin · Accepted Answer

For the reaction, CO 2(g)+ C (s) leftharpoons 2 CO (g) Given, KC=0.036 Equilibrium conc. of [ CO 2(g)]=0.004 M As we know that, KC=([ CO (g)]2/[ CO 2(g)]) [ C (s)=1] ldots (i) On substituting the given values in eq. (i), we get, 0.036 =([ CO (g)]2/0.004) [ CO (g)] =√0.036 × 0.004 =√1.44 × 10-4 =1.2 × 10-2 mol / L

Q. The value of $K_{C}$ for the equilibrium reaction
$CO X_{2 (g)} + C X_{(s)} 2 CO X_{(g)}$
at $T (K)$ is $0.036.$ If the equilibrium concentration of $C O_{2} (g)$ is $0.004 M$ , the concentration of $CO (g)$ in $m o l L^{- 1}$ is

$3.6 \times 1 0^{- 2}$

$2.0 \times 1 0^{- 2}$

$1.2 \times 1 0^{- 2}$

$1.2 \times 1 0^{- 3}$

Q. The value of KC​ for the equilibrium reaction COX2(g)​​+CX(s)​​​2COX(g)​​ at T(K) is 0.036. If the equilibrium concentration of CO2​(g) is 0.004M, the concentration of CO(g) in molL−1 is

3.6×10−2

2.0×10−2

1.2×10−2

1.2×10−3

For the reaction, CO2​(g)+C(s)⇌2CO(g) Given, KC​=0.036 Equilibrium conc. of [CO2​(g)]=0.004M As we know that, KC​=[CO2​(g)][CO(g)]2​[C(s)=1]… (i) On substituting the given values in eq. (i), we get, 0.036=0.004[CO(g)]2​ [CO(g)]=0.036×0.004​ =1.44×10−4​ =1.2×10−2mol/L

Q. The value of $K_{C}$ for the equilibrium reaction
$CO X_{2 (g)} + C X_{(s)} 2 CO X_{(g)}$
at $T (K)$ is $0.036.$ If the equilibrium concentration of $C O_{2} (g)$ is $0.004 M$ , the concentration of $CO (g)$ in $m o l L^{- 1}$ is

$3.6 \times 1 0^{- 2}$

$2.0 \times 1 0^{- 2}$

$1.2 \times 1 0^{- 2}$

$1.2 \times 1 0^{- 3}$