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What will be the equilibrium constant at ${127^o}C$ if equilibrium constant at ${27^o}C$ is $4$ for the reaction ${{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}};\,\,\,\Delta H=-46.06kJ$ .(A) $4 \times {10^{ - 2}}$ (B) $2 \times {10^{ - 3}}$ (C) ${10^2}$ (D) $4 \times {10^2}$

Last updated date: 15th Jul 2024
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Hint :When the rate of forward reaction balances the rate of backward reaction and there is no tendency for any further change, then the system is said to be in equilibrium. The value of reaction quotient at equilibrium state is termed as equilibrium constant.

Van’t Hoff equation: This is an equation which relates equilibrium constant of the system to the enthalpy of reaction at a given temperature. The expression is given as follows:
$\dfrac{d}{{dT}}\left[ {\ln {K_{eq}}} \right] = \dfrac{{\Delta H}}{{R{T^2}}}\,\,\,\, - (i)$
Where, ${K_{eq}}$ is the equilibrium constant, $\Delta H$ is the change in enthalpy of the reaction, $R$ is the universal gas constant and $T$ is the temperature at which equilibrium is achieved.
Integrating equation $(i)$ between temperature ${T_1}$ and ${T_2}$ :
$\int\limits_{{K_1}}^{{K_2}} {d\left[ {\ln {K_{eq}}} \right]} = \dfrac{{\Delta H}}{R}\int\limits_{{T_1}}^{{T_2}} {\dfrac{1}{{{T^2}}}dT} \,$
$\Rightarrow \ln \left( {\dfrac{{{K_2}}}{{{K_1}}}} \right) = \dfrac{{\Delta H}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)\,\, - (ii)$
Now, data given in the question is as follows:
${K_1} = 4$
${T_1} = {27^o}C$
$\Rightarrow 27 + 273 = 300{\rm{ K}}$
${T_2} = {127^o}C$
$\Rightarrow 127 + 273 = 400{\rm{ K}}$
Change in enthalpy $\Delta H = - 46.06kJ$
$\Rightarrow \Delta H = - 46.06 \times {10^3}J$
We need to find the equilibrium constant ${K_2}$ at temperature ${T_2}$ . Substituting values in equation $(ii)$ :
$\ln \left( {\dfrac{{{K_2}}}{4}} \right) = \dfrac{{ - 46.06 \times {{10}^3}}}{{8.314}}\left( {\dfrac{1}{{300}} - \dfrac{1}{{400}}} \right)$
$\Rightarrow \ln \left( {\dfrac{{{K_2}}}{4}} \right) = - 5.54 \times {10^3}\left( {\dfrac{{400 - 300}}{{300 \times 400}}} \right)$
$\Rightarrow \ln \left( {\dfrac{{{K_2}}}{4}} \right) = - 4.62$
Taking antilog on both sides of the equation:
$\Rightarrow \dfrac{{{K_2}}}{4} = {e^{ - 4.62}}$
$\Rightarrow {K_2} = 0.04$
$\Rightarrow {K_2} = 4 \times {10^{ - 2}}$
Hence, the equilibrium constant for the given reaction at ${127^o}C$ is $4 \times {10^{ - 2}}$ .
Thus, option (A) is the correct answer.

Note :
Ensure to check the units of the given data before substituting it in the expression. The temperature considered here is absolute temperature i.e., we need to convert the given values of temperature in Kelvin. Also, the value of universal gas constant is always to be decided on the basis of the other given variables, which means enthalpy of the reaction is given in units of joule, so the value of $R$ must also be expressed in Joule.