Question

consider the following system at equilibrium where \\( \delta h = -10.4 \mathrm{kj} \\), and \\( k_{c} = 55.6 \\), at \\( 6.98 \times 10^{2} \mathrm{k} \\).
\\( \mathrm{h}_{2}(g)+\mathrm{i}_{2}(g) \
ightleftharpoons 2 \mathrm{hi}(g) \\)
if the temperature on the equilibrium system is suddenly decreased:
the value of \\( k_{c} \\)
oincreases
odecreases
oremains the same
the reaction must
orun in the forward direction to reestablish equilibrium
orun in the reverse direction to reestablish equilibrium
oremain in the current position, since it is already at equilibrium
the concentration of \\( \mathrm{i}_{2} \\) will
oincrease
odecrease
oremain the same

Explanation:

Step1: Analyze the effect of temperature change on equilibrium constant

For an exothermic reaction ($\Delta H=- 10.4\ \text{kJ}$), when the temperature is decreased, according to Le - Chatelier's principle, the equilibrium will shift in the forward direction.

Step2: Determine the change in $K_{c}$

The equilibrium constant $K_{c}$ is a measure of the ratio of products to reactants at equilibrium. For an exothermic reaction, a decrease in temperature favors the forward reaction. As the forward reaction is favored, the concentration of products (in this case, $HI$) increases relative to reactants ($H_{2}$ and $I_{2}$). The formula for $K_{c}=\frac{[HI]^{2}}{[H_{2}][I_{2}]}$. When the forward reaction is favored (more $HI$ is formed), the value of $K_{c}$ increases.

Answer:

Increases