Question

For the reaction, ${\text{CO}}\left( {\text{g}} \right){\text{ + 2}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right){\text{ }} \rightleftharpoons {\text{ C}}{{\text{H}}_{\text{3}}}{\text{OH}}\left( {\text{g}} \right)$ , hydrogen gas is introduced into a $5{\text{ L}}$ flask at $327^\circ {\text{C}}$ , containing $0.2{\text{ mole}}$ of ${\text{CO}}\left( {\text{g}} \right)$ and a catalyst (solid), until the pressure is ${\text{4}}{\text{.92 atmosphere}}$ . At this point, ${\text{0}}{\text{.1 mole}}$ of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\left( {\text{g}} \right)$ is formed. Calculate the equilibrium constants ${K_P}$ and ${K_C}$.

Answer 1

To solve the above question we need to firstly use the ideal gas equation ${\text{P}} \times {\text{V = n}} \times {\text{R}} \times {\text{T}}$ to calculate the number of moles. After finding the number of moles use the expression ${K_P} = {K_C} \times {\left( {RT} \right)^{\Delta n}}$ to calculate ${K_P}$ from ${K_C}$

Complete Step by step answer: Let ${\text{m moles}}$ of hydrogen are added to the flask. The total number of moles of carbon monoxide, methanol and hydrogen will also be ${\text{m moles}}$.
Write the ideal gas equation
${\text{P}} \times {\text{V = n}} \times {\text{R}} \times {\text{T}}$
Here, P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant and T is the absolute temperature.
Substitute ${\text{4}}{\text{.92 atmosphere}}$ for P, $5{\text{ L}}$for V, 0.082 for R and $327 + 273 = 600{\text{ K}}$ for T in the ideal gas equation and calculate the number of moles
${\text{4}}{\text{.92 atmosphere}} \times 5{\text{ L = m}} \times {\text{0}}{\text{.082}} \times {\text{600K}}$
${\text{m}} = \dfrac{{{\text{4}}{\text{.92 atmosphere}} \times 5{\text{ L}}}}{{{\text{0}}{\text{.082}} \times {\text{600K}}}}$
${\text{m}} = 0.5{\text{ mole}}$
Let x moles of methanol are formed at equilibrium

| ${\text{CO}}$| ${{\text{H}}_{\text{2}}}$| ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$
---|---|---|---
Number of moles| $0.2 - x$| $y - 2x$| $x$
Number of moles| $0.2 - 0.1 = 0.1$| $0.3 - 2\left( {0.1} \right) = 0.1$| $0.1$
Active mass| $\dfrac{{0.1}}{5}$| $\dfrac{{0.1}}{5}$| $\dfrac{{0.1}}{5}$

Write the equilibrium constant expression
${K_C} = \dfrac{{\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}} \right]}}{{\left[ {{\text{CO}}} \right] \times {{\left[ {{{\text{H}}_{\text{2}}}} \right]}^2}}}$
Substitute values in the above expression and calculate the value of the equilibrium constant ${K_C}$.
${K_C} = \dfrac{{\dfrac{{0.1}}{5}}}{{\dfrac{{0.1}}{5} \times {{\left( {\dfrac{{0.1}}{5}} \right)}^2}}}$
$\Rightarrow {K_C} = {\left( {\dfrac{5}{{0.1}}} \right)^2}$
$\Rightarrow {K_C} = {\left( {50} \right)^2}$
$\Rightarrow {K_C} = 2500$
Hence, the value of the equilibrium constant ${K_C}$is 2500.
Write the expression between ${K_C}$ and ${K_P}$
${K_P} = {K_C} \times {\left( {RT} \right)^{\Delta n}}$
Substitute values in the above expression and calculate ${K_P}$
$\Rightarrow {K_P} = 2500 \times {\left( {0.082 \times 600} \right)^{ - 2}}$
$\Rightarrow {K_P} = \dfrac{{2500}}{{{{\left( {49.2} \right)}^2}}}$
$\Rightarrow{K_P} = 1.0327$

Hence, the value of the equilibrium constant ${K_P}$ is 1.0327.

Note: $\Delta n$ represents the difference between the number of moles of gaseous products and the number of moles of gaseous reactants. In the reaction between carbon monoxide and hydrogen to form methanol, one mole of gaseous product and three moles of gaseous reactants are present.
$\Delta n = 1 - \left( {1 + 2} \right) = - 2$