For the reaction Cl_{2(g)} + 3F_{2(g)} rightl | vedclass

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(C) The given reaction is $Cl_{2(g)} + 3F_{2(g)} \rightleftharpoons 2ClF_{3(g)}$ with $\Delta H = -329 \, kJ$. Since $\Delta H < 0$,the reaction is ex

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Adding $F_2$

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(C) The given reaction is $Cl_{2(g)} + 3F_{2(g)} \rightleftharpoons 2ClF_{3(g)}$ with $\Delta H = -329 \, kJ$. Since $\Delta H According to Le Chatelier's principle,for an exothermic reaction,increasing the temperature shifts the equilibrium to the left (reactant side),decreasing the yield of $ClF_3$. Adding more reactant $(F_2)$ shifts the equilibrium to the right (product side),increasing the amount of $ClF_3$. Removing $Cl_2$ shifts the equilibrium to the left,decreasing the yield. Increasing the volume of the container favors the side with more moles of gas. Here,reactants have $1 + 3 = 4$ moles and products have $2$ moles. Increasing volume shifts the equilibrium to the left,decreasing the yield of $ClF_3$.

For the reaction $Cl_{2(g)} + 3F_{2(g)} \rightleftharpoons 2ClF_{3(g)}; \Delta H = -329 \, kJ$,which of the following will increase the amount of $ClF_{3(g)}$ in the equilibrium mixture?

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The exothermic formation of $ClF_3$ is represented by the equation $Cl_{2(g)} + 3F_{2(g)} \rightleftharpoons 2ClF_{3(g)}$; $\Delta H = -329 \ kJ$. 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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