For the reaction FeO(s)+CO(g)Fe(s)+CO2(g), KC=5.0. If the concentration of CO at equilibrium is 2.5× 10-2mol/litre at 100oC . Then the equilibrium concentration of CO2 will be:

Question

For the reaction FeO(s)+CO(g)Fe(s)+CO2(g), KC=5.0. If the concentration of CO at equilibrium is 2.5× 10-2mol/litre at 100oC . Then the equilibrium concentration of CO2 will be: (A)  1.25× 10-4m/L  (B)  12.5× 10-2m/L  (C)  2.50× 10-3m/L  (D)  2.50× 10-4m/L

Tardigrade Admin · Accepted Answer

FeO(s)+CO(g) xrightarrowFe(s)+CO2(g) At equilibrium KC=([CO2]/[CO]) given [CO]=2.5× 10-2mol/litre, KC=5.0 5=([CO2]/2.5× 10-2) [CO2]=12.5× 10-2mol/litre

Q. For the reaction $F e O_{(s)} + C O_{(g)} F e_{(s)} + C O_{2}_{(g)}, K_{C} = 5.0.$ If the concentration of CO at equilibrium is $2.5 \times 10^{- 2} m o l / l i t re$ at $100^{o} C$ . Then the equilibrium concentration of $C O_{2}$ will be:

$1.25 \times 10^{- 4} m / L$

$12.5 \times 10^{- 2} m / L$

$2.50 \times 10^{- 3} m / L$

$2.50 \times 10^{- 4} m / L$

$F e O_{(s)} + C O_{(g)} F e_{(s)} + C O_{2 (g)}$ At equilibrium $K_{C} = \frac{[ C O _{2} ]}{[ CO ]}$ given $[CO] = 2.5 \times 10^{- 2} m o l / l i t re, K_{C} = 5.0$ $5 = \frac{[ C O _{2} ]}{2.5 \times 10 ^{- 2}}$ $[C O_{2}] = 12.5 \times 10^{- 2} m o l / l i t re$

Q. For the reaction FeO(s)​+CO(g)​Fe(s)​+CO2​(g)​,KC​=5.0. If the concentration of CO at equilibrium is 2.5×10−2mol/litre at 100oC . Then the equilibrium concentration of CO2​ will be:

1.25×10−4m/L

12.5×10−2m/L

2.50×10−3m/L

2.50×10−4m/L