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Use the reaction [tex]I_2(s) \rightleftharpoons I_2(g), \Delta H=62.4 \, \text{kJ/mol}, \Delta S=0.145 \, \text{kJ/(mol \cdot K)}[/tex] for this question.

At what temperature is the reaction at equilibrium?

A. [tex]0.002 \, \text{K}[/tex]
B. [tex]62 \, \text{K}[/tex]
C. [tex]157 \, \text{K}[/tex]
D. [tex]430 \, \text{K}[/tex]

Sagot :

GinnyAnswer
To determine the temperature at which the reaction [tex]\( I_2(s) \rightleftharpoons I_2(g) \)[/tex] is at equilibrium, we start with the concept of Gibbs free energy [tex]\(\Delta G\)[/tex]. At equilibrium, [tex]\(\Delta G\)[/tex] is zero. The relationship between [tex]\(\Delta G\)[/tex], [tex]\(\Delta H\)[/tex], and [tex]\(\Delta S\)[/tex] is given by the equation:

[tex]\[ \Delta G = \Delta H - T\Delta S \][/tex]

At equilibrium, we set [tex]\(\Delta G\)[/tex] to zero:

[tex]\[ 0 = \Delta H - T\Delta S \][/tex]

We need to solve this equation for [tex]\( T \)[/tex]:

[tex]\[ T = \frac{\Delta H}{\Delta S} \][/tex]

We are given:
[tex]\[ \Delta H = 62.4 \text{ kJ/mol} \][/tex]
[tex]\[ \Delta S = 0.145 \text{ kJ/(mol*K)} \][/tex]

Now we plug these values into the equation:

[tex]\[ T = \frac{62.4 \text{ kJ/mol}}{0.145 \text{ kJ/(mol*K)}} \][/tex]

Performing the division gives us:

[tex]\[ T = 430.3448275862069 \text{ K} \][/tex]

Upon examining the options provided:

A. [tex]\(0.002 \text{ K}\)[/tex]
B. [tex]\(62 \text{ K}\)[/tex]
C. [tex]\(157 \text{ K}\)[/tex]
D. [tex]\(430 \text{ K}\)[/tex]

The closest value to our calculated temperature is [tex]\(430 \text{ K}\)[/tex]. Therefore, the correct answer is:

D. [tex]\(430 \text{ K}\)[/tex]
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