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 moles of $C_{2} H_{4}$ can be increased by:

increasing the temperature

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decreasing the temperature

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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 options are
A increasing the temperature
C removing some $H_{2}$
D adding some $C_{2} H_{6}$
Since $Δ H$ is negative, it is an exothermic reaction. Hence, by using Le Chatelier's principle, increasing the temperature will favor the formation of $C_{2} H_{4}$ . Also, the same principle can be applied for decreasing the concentration of $H_{2}$ which means removing some $H_{2}$ and increasing the concentration of $C_{2} H_{6}$ or by adding $C_{2} H_{6}$ into the closed vessel.

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For the gas phase reaction, C2H4+H2⇌C2H6(ΔH=−32.7kcal), carried out in a closed vessel, the equilibrium moles of C2H4 can be increased by:

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 moles of $C_{2} H_{4}$ can be increased by: