The exothermic formation of ClF_3 is represen | vedclass

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(D) The reaction is exothermic $(\Delta H_r < 0)$. According to Le Chatelier's principle: $1$. Increasing the temperature shifts the equilibrium to th

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Decreasing the volume of the container

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(D) The reaction is exothermic $(\Delta H_r $1$. Increasing the temperature shifts the equilibrium to the left (reactant side). $2$. Adding an inert gas at constant volume has no effect on the equilibrium position. $3$. Removing $Cl_2$ shifts the equilibrium to the left. $4$. Decreasing the volume of the container increases the pressure,shifting the equilibrium toward the side with fewer moles of gas. In this reaction,there are $4$ moles of gas on the reactant side and $2$ moles on the product side. Thus,decreasing the volume shifts the equilibrium to the right,increasing the yield of $ClF_3$.

The exothermic formation of $ClF_3$ is represented by the equation
$Cl_{2(g)} + 3F_{2(g)} \rightleftharpoons 2ClF_{3(g)}$; $\Delta H_r = -329 \ kJ$
Which of the following will increase the quantity of $ClF_3$ in an equilibrium mixture of $Cl_2, F_2$ and $ClF_3$?

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The exothermic formation of $ClF_3$ is represented by the equation$Cl_{2(g)} + 3F_{2(g)} \rightleftharpoons 2ClF_{3(g)}$; $\Delta H_r = -329 \ kJ$Which of the following will increase the quantity of $ClF_3$ in an equilibrium mixture of $Cl_2, F_2$ and $ClF_3$?