The equilibrium constant kp for the reaction H2(g)+CO2(g)⇌H2O(g)+CO(g)is 16.0 at 1600∘C. Initially 0.80 mole H2 and 0.80 mole CO2 are injected into a 5.0 lit flask. What is the equilibrium concentration of CO2(g)?

H2(g)+CO2(g)⇌H2O(g)+CO(g) At eq. 0.8-x 0.8-x x xConc at eq. 0.8−x5 0.8−x5 x5 x5 ∴Δn(g)=o kP=kc ∴kc=(x/5)2(0.8−x5)2 ∴x=3.25=0.64[CO2(g)]=0.8−0.645=0.032

The equilibrium constant kp for the reaction H2(g)+CO2(g)⇌H2O(g)+CO(g)is 16.0 at 1600∘C. Initially 0.80 mole H2 and 0.80 mole CO2 are injected into a 5.0 lit flask. What is the equilibrium concentration of CO2(g)?

H2(g)+CO2(g)⇌H2O(g)+CO(g) At eq. 0.8-x 0.8-x x xConc at eq. 0.8−x5 0.8−x5 x5 x5 ∴Δn(g)=o kP=kc ∴kc=(x/5)2(0.8−x5)2 ∴x=3.25=0.64[CO2(g)]=0.8−0.645=0.032

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The equilibrium constant $k_{p}$ for the reaction
$H_{2 (g)} + C O_{2 (g)} ⇌ H_{2} O_{(g)} + C O_{(g)}$
is 16.0 at $1600^{\circ} C$ . Initially 0.80 mole $H_{2}$ and 0.80 mole $C O_{2}$ are injected into a 5.0 lit flask. What is the equilibrium concentration of $C O_{2 (g)} ?$

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0.035

0.034

0.033

0.032

answer is B.

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Detailed Solution

$H_{2 (g)} + C O_{2 (g)} ⇌ H_{2} O_{(g)} + C O_{(g)}$

At eq. 0.8-x 0.8-x x x
Conc at eq. $\frac{0.8 - x}{5}$    $\frac{0.8 - x}{5}$ $\frac{x}{5}$ $\frac{x}{5}$
$∴ Δ n_{(g)} = o k_{P} = k_{c}$    $∴ k_{c} = \frac{{(x / 5)}^{2}}{{(\frac{0.8 - x}{5})}^{2}}$

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$∴ x = \frac{3.2}{5} = 0.64$
$[C O_{2 (g)}] = \frac{0.8 - 0.64}{5} = 0.032$