For the gas phase reaction, $C_{2} H_{4} + H_{2}$ $⇌$ $C_{2} H_{6}$ $(Δ H = - 32.7 k c a l)$ , carried out in a closed vessel. The equilibrium concentration of $C_{2} H_{4}$ can be increased by?

increasing the temperature

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decreasing the pressure

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removing some

H_{2}

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adding some

C_{2} H_{6}

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Solution

The correct option is C increasing the temperature
$C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6}$ $Δ H = - 32.7 K C a l$
$K_{e q b^{m}} = \frac{[C_{2} H_{6}]}{[C_{2} H_{4}] [H_{2}]} = K_{1}$
After increased in concentration of $C_{2} H_{4}$
$\Rightarrow K_{2} = \frac{[C_{2} H_{6}]^{1}}{[C_{2} H_{4}]^{1} [H_{2}]^{1}}$
$K_{2} < K_{1}$
Van't Hoff equation
$\Rightarrow l n \frac{K_{2}}{K_{1}} = - \frac{Δ H}{R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$
as $K_{2} < K_{1}$ , thus $l n \frac{K_{2}}{K_{1}}$ has negative value.
as $Δ H$ is exothermic and has negative value.
To make $l n \frac{K_{2}}{K_{1}}$ negative, $T_{2}$ must be greater than $T_{1}$ .
Thus, to increase the concentration of $C_{2} H_{4}$ we have to increase the temperature.

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For the gas phase reaction $C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6},$ $Δ H$ = -32.7kcal carried out in a vessel, the equilibrium concentration of $C_{2} H_{4}$ can be increased by -

For the gas phase reaction $C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6},$ DH = -32.7kcal carried out in a vessel, the equilibrium concentration of $C_{2} H_{4}$ can be increased by -