For the gas phase reaction- C 2 H 4 g - H 2 g - C 2 H 6 g - H 0-136-8 kJ mol -1carried out in a vessel- the equilibrium concentration of C 2 H 4 can be increased by-A- Increasing the temperatureB- Decreasing the pressureC- Removing some H 2D- Adding some C 2 H 6

The correct options are A Increasing the temperature nbsp; B Decreasing the pressure nbsp; C Removing some H_2 D Adding some C_2H_6a. As the reaction is exo ...

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Byju's Answer

Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

For the gas p...

Question

For the gas phase reaction,
$C_{2} H_{4} (g) + H_{2} (g) ⇌ C_{2} H_{6} (g); △ H^{o} = - 136.8 k J m o l^{- 1}$
carried out in a 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 Decreasing the pressure
C Removing some $H_{2}$
D Adding some $C_{2} H_{6}$
a. As the reaction is exothermic, the temperature reaction will shift in a backward direction, i.e., concentration of $C_{2} H_{4}$ will increase.

b. The reaction is carried out in the gas phase, hence on decreasing the pressure the reaction will shift in the direction where there are more number of moles.
Hence, reaction will shift on the reactant's side.

c. Removing the reactant, $H_{2}$ will shift the equilibrium to the backward direction and hence concentration of $C_{2} H_{4}$ will increase.

d. Adding the product $C_{2} H_{6}$ will shift the equilibrium to the backward direction and hence concentration of $C_{2} H_{4}$ will increase.

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For the gas phase reaction, C2H4(g)+H2(g)⇌C2H6(g); △Ho=−136.8 kJ mol−1 carried out in a vessel, the equilibrium concentration of C2H4 can be increased by:

For the gas phase reaction,
$C_{2} H_{4} (g) + H_{2} (g) ⇌ C_{2} H_{6} (g); △ H^{o} = - 136.8 k J m o l^{- 1}$
carried out in a vessel, the equilibrium concentration of $C_{2} H_{4}$ can be increased by: