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Le Chatelier's Principle for Del N Equals to 0

For the gas p...

Question

For the gas phase reaction, $C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6} [Δ H^{o} = - 136.8 k J {m o l}^{- 1}]$ , is carried out in 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 options are
A increasing the temperature
B removing some $H_{2}$
C decreasing the pressure
D adding some $C_{2} H_{6}$
The given reaction is :-
$C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6}; {△ H}^{0} = - 136.8 K J / m o l$
(I) This reaction is exothermic reaction, so increase of temperature favours backward reaction and formation of $C_{2} H_{6}$ is lowered.
(II) High pressure favours the reaction which takes place with decrease in no. of moles. So, an decreasing the pressure, the formation of $C_{2} H_{6}$ is decreased.
(III) Removing of $H_{2}$ increases backward reaction rate as per Le Chatelier's principle.
(IV) Adding of $C_{2} H_{6}$ also decreases the forward reaction as predicted by Le Chatelier's principle.

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