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# Formation of $Cl{F_3}$ from $C{l_2}$ and ${F_2}$ is an exothermic process. The equilibrium system can be represented as:$C{l_{2\left( g \right)}} + 3{F_2} \rightleftharpoons 2Cl{F_3}$ ; $\Delta H = - 329kJ$Which of the following will increase the quantity of $Cl{F_3}$ in the equilibrium mixture?A) Increase in temperature, decrease in pressure, addition of $C{l_2}$ .B) Decrease in temperature and pressure, addition of $Cl{F_3}$ .C) Increase in temperature and pressure removal of $C{l_2}$ .D) Decrease in temperature, increase in pressure, addition of ${F_2}$ .

Last updated date: 13th Jun 2024
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Hint:We Know If a chemical system at equilibrium experiences a change in, temperature volume, concentration, or partial pressure, at that point the equilibrium shifts to neutralize the imposed change and a new equilibrium is recognized.
Example:
$2{H_2}\left( g \right) + {O_2}\left( g \right) \to 2{H_2}O\left( g \right)$
Increasing the concentration of reactants shifts the equilibrium to the right side and thus concentration of the water also increases.

The given reaction is,
$C{l_{2\left( g \right)}} + 3{F_2} \rightleftharpoons 2Cl{F_3}$ ; $\Delta H = - 329kJ$
We have to know that as it is releases energy, it is an exothermic response decline in temperature favors forward response, an expansion in the centralization of reactants favors forward response and an increment in weight in the weight additionally favors as $\Delta {n_g} < 0$ .Exothermic responses favors creation of bonds i.e. development of $Cl{F_3}$, and the response continues towards right side at low temperatures, high weights and furthermore the option of reactants fluorine.
Therefore the correct option is option (D).

Note:
-We have to know that the equilibrium expression for benzoic acid is,
${C_6}{H_5}COOH\left( {aq} \right) + {H_2}O\left( l \right) \to {H_3}{O^ + }\left( {aq} \right) + {C_6}{H_5}C{O_2}^ -$
-Adding hydrochloric acid will increase the hydronium ion concentration of the acetate ion so the reaction will shift to the left to achieve equilibrium causing increase in the benzoic acid concentration and the percentage protonation of benzoic acid gets decreased.
-The given reaction is,
${H_2}\left( g \right) + {F_2}\left( g \right) \rightleftarrows 2HF\left( g \right)$
-According to Le-Chatteliers principle increasing the concentration of Hydrogen fluoride shifts the equilibrium to the left to form more reactants.
-The given reaction is,
${{\text{H}}_{\text{2}}}\left( {\text{g}} \right){\text{ + }}{{\text{I}}_{\text{2}}}\left( {\text{g}} \right) \rightleftarrows {\text{2HI}}\left( {\text{g}} \right)$
-According to Le-Chatelier's principle decreasing or increasing the concentration hydrogen iodide shifts the equilibrium to the right to form more products. While decreasing or increasing the concentration of iodine shifts the equilibrium to the left to form more reactants.