In which of the following reactions at equili | vedclass

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(C) According to Le Chatelier's principle,when the total pressure of a system at equilibrium is increased,the equilibrium shifts in the direction that

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$(I)$

Vedclass · Answer

(C) According to Le Chatelier's principle,when the total pressure of a system at equilibrium is increased,the equilibrium shifts in the direction that results in a decrease in the total number of gaseous moles.For reaction $(I)$: $X_{2(g)} + 3Y_{2(g)} \rightleftharpoons 2XY_{3(g)}$. The number of gaseous moles on the reactant side is $1 + 3 = 4$,and on the product side is $2$. Since $2 For reactions $(II)$,$(III)$,and $(IV)$: The number of gaseous moles on the reactant side is $1 + 1 = 2$,and on the product side is $2$. Since the number of moles is equal on both sides,a change in pressure does not affect the position of the equilibrium.

Similar Questions

The $\%$ yield of ammonia as a function of time in the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,$\Delta H < 0$ at $(P, T_1)$ is given below. If this reaction is conducted at $(P, T_2)$,with $T_2 > T_1$,the $\%$ yield of ammonia as a function of time is represented by:

$aA \rightleftharpoons mB + nC$,$\Delta H = + x \ kcal$. If $m + n < a$,then conditions favouring forward reaction will be

$2x_{(g)} + y_{(g)} \rightleftarrows 2z_{(g)} + 80 \ kcal$
Which combination gives the highest yield of $z$ at equilibrium?

For which of the following reactions is product formation favoured at low pressure and low temperature?

For the formation of $NH_{3(g)}$ from $H_{2(g)}$ and $N_{2(g)}$,which one of the following graphs represents the yield of ammonia at equilibrium? (Assume $T_1 < T_2$,unless mentioned)

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