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The equilibrium constant, [tex]$K _{ p }$[/tex], for the following reaction [tex]$H_2(g) + Cl_2(g) \rightarrow 2 HCl (g)$[/tex] at [tex]25^{\circ} C[/tex] is 13.75. Which of the following conditions are NOT at equilibrium?

Select one:

a. [tex]P_{H_2} = 0.085 \, \text{atm}, \, P_{Cl_2} = 0.085 \, \text{atm}, \, P_{HCl} = 0.32 \, \text{atm}[/tex]

b. [tex]P_{H_2} = 0.270 \, \text{atm}, \, P_{Cl_2} = 0.270 \, \text{atm}, \, P_{HCl} = 1 \, \text{atm}[/tex]

c. [tex]P_{H_2} = 1.25 \, \text{atm}, \, P_{Cl_2} = 0.121 \, \text{atm}, \, P_{HCl} = 0.5 \, \text{atm}[/tex]

d. [tex]P_{H_2} = 0.125 \, \text{atm}, \, P_{Cl_2} = 1.31 \, \text{atm}, \, P_{HCl} = 1.5 \, \text{atm}[/tex]

e. All of the above are at equilibrium

Sagot :

GinnyAnsweridn
Let's determine which of the given conditions are not at equilibrium for the reaction:

[tex]\[ \text{H}_2(g) + \text{Cl}_2(g) \leftrightarrows 2 \text{HCl}(g) \][/tex]

The equilibrium constant, [tex]\( K_p \)[/tex], at [tex]\( 25^\circ C \)[/tex] is 13.75. The reaction quotient, [tex]\( Q_p \)[/tex], is given by:

[tex]\[ Q_p = \frac{(P_{\text{HCl}})^2}{P_{\text{H}_2} \cdot P_{\text{Cl}_2}} \][/tex]

We will calculate [tex]\( Q_p \)[/tex] for each condition and compare it to [tex]\( K_p \)[/tex].

Condition a:
[tex]\[ P_{\text{H}_2} = 0.085 \, \text{atm}, \, P_{\text{Cl}_2} = 0.085 \, \text{atm}, \, P_{\text{HCl}} = 0.32 \, \text{atm} \][/tex]

[tex]\[ Q_p = \frac{(0.32)^2}{0.085 \cdot 0.085} = \frac{0.1024}{0.007225} \approx 14.18 \][/tex]

Condition b:
[tex]\[ P_{\text{H}_2} = 0.270 \, \text{atm}, \, P_{\text{Cl}_2} = 0.270 \, \text{atm}, \, P_{\text{HCl}} = 1.0 \, \text{atm} \][/tex]

[tex]\[ Q_p = \frac{(1.0)^2}{0.270 \cdot 0.270} = \frac{1.0}{0.0729} \approx 13.72 \][/tex]

Condition c:
[tex]\[ P_{\text{H}_2} = 1.25 \, \text{atm}, \, P_{\text{Cl}_2} = 0.121 \, \text{atm}, \, P_{\text{HCl}} = 0.5 \, \text{atm} \][/tex]

[tex]\[ Q_p = \frac{(0.5)^2}{1.25 \cdot 0.121} = \frac{0.25}{0.15125} \approx 1.65 \][/tex]

Condition d:
[tex]\[ P_{\text{H}_2} = 0.125 \, \text{atm}, \, P_{\text{Cl}_2} = 1.31 \, \text{atm}, \, P_{\text{HCl}} = 1.5 \, \text{atm} \][/tex]

[tex]\[ Q_p = \frac{(1.5)^2}{0.125 \cdot 1.31} = \frac{2.25}{0.16375} \approx 13.74 \][/tex]

Now we compare [tex]\( Q_p \)[/tex] to [tex]\( K_p = 13.75 \)[/tex]. We find:

- For condition a, [tex]\( Q_p \approx 14.18 \ne 13.75 \)[/tex]. Thus, it is not at equilibrium.
- For condition b, [tex]\( Q_p \approx 13.72 \ne 13.75 \)[/tex]. Thus, it is not at equilibrium.
- For condition c, [tex]\( Q_p \approx 1.65 \ne 13.75 \)[/tex]. Thus, it is not at equilibrium.
- For condition d, [tex]\( Q_p \approx 13.74 \ne 13.75 \)[/tex]. Thus, it is not at equilibrium.

Since all the calculated [tex]\( Q_p \)[/tex] values are not equal to [tex]\( K_p \)[/tex], all of the above conditions are not at equilibrium.

The correct selection is:
[tex]\[ \boxed{\text{e. all of the above are at equilibrium}} \][/tex]
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