The reaction C_2H_{6(g)} rightleftharpoons C_ | vedclass

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(D) $C_2H_{6(g)} \rightleftharpoons C_2H_{4(g)} + H_{2(g)}$,$\Delta H = 137.0 \ kJ \ mol^{-1}$Increasing the volume of the closed reaction vessel woul

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Increasing the volume of the closed reaction vessel

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(D) $C_2H_{6(g)} \rightleftharpoons C_2H_{4(g)} + H_{2(g)}$,$\Delta H = 137.0 \ kJ \ mol^{-1}$Increasing the volume of the closed reaction vessel would shift the equilibrium to the right.As the volume increases,pressure decreases,so the system shifts to increase the number of gaseous molecules.In the given reaction,the number of moles of gaseous products $(2 \ mol)$ is greater than the number of moles of gaseous reactants $(1 \ mol)$.Thus,according to Le Chatelier's principle,the equilibrium will shift in the forward direction to increase the number of gaseous molecules.

The reaction $C_2H_{6(g)} \rightleftharpoons C_2H_{4(g)} + H_{2(g)}$ is at equilibrium in a closed vessel at $1000 \ K$. The enthalpy change $(\Delta H)$ for the reaction is $137.0 \ kJ \ mol^{-1}$. Which one of the following actions would shift the equilibrium to the right?

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