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 ${F_2}$ .

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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.
$2{H_2}\left( g \right) + {O_2}\left( g \right) \to 2{H_2}O\left( g \right)$
Increasing the concentration of reactants shifts the equilibrium to the right side and thus concentration of the water also increases.

Complete step by step answer:
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).

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${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}^ - $
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