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To determine if the conditions are at equilibrium for the reaction [tex]\(H_2(g) + Cl_2(g) \rightleftharpoons 2 HCl(g)\)[/tex], where the equilibrium constant [tex]\(K_p\)[/tex] is 13.75 at 25°C, we can compare the reaction quotient [tex]\(Q\)[/tex] for each set of initial conditions to the equilibrium constant [tex]\(K_p\)[/tex]. The reaction quotient [tex]\(Q\)[/tex] is calculated using the same formula used for the equilibrium constant but with the current concentrations (or partial pressures if we were dealing with gases).
The reaction quotient [tex]\(Q\)[/tex] for this reaction is given by:
[tex]\[ Q = \frac{[HCl]^2}{[H_2] \cdot [Cl_2]} \][/tex]
Now, let's calculate [tex]\(Q\)[/tex] for each set of initial conditions and compare it with [tex]\(K_p = 13.75\)[/tex].
### Initial Conditions and Calculations
1. Condition 1:
- [tex]\( [H_2] = 0.1 \)[/tex]
- [tex]\( [Cl_2] = 0.2 \)[/tex]
- [tex]\( [HCl] = 1.5 \)[/tex]
[tex]\[ Q = \frac{(1.5)^2}{(0.1) \cdot (0.2)} = \frac{2.25}{0.02} = 112.50 \][/tex]
2. Condition 2:
- [tex]\( [H_2] = 0.5 \)[/tex]
- [tex]\( [Cl_2] = 0.5 \)[/tex]
- [tex]\( [HCl] = 2 \)[/tex]
[tex]\[ Q = \frac{(2)^2}{(0.5) \cdot (0.5)} = \frac{4}{0.25} = 16.00 \][/tex]
3. Condition 3:
- [tex]\( [H_2] = 0.3 \)[/tex]
- [tex]\( [Cl_2] = 0.2 \)[/tex]
- [tex]\( [HCl] = 1 \)[/tex]
[tex]\[ Q = \frac{(1)^2}{(0.3) \cdot (0.2)} = \frac{1}{0.06} = 16.67 \][/tex]
4. Condition 4:
- [tex]\( [H_2] = 0.4 \)[/tex]
- [tex]\( [Cl_2] = 0.4 \)[/tex]
- [tex]\( [HCl] = 4 \)[/tex]
[tex]\[ Q = \frac{(4)^2}{(0.4) \cdot (0.4)} = \frac{16}{0.16} = 100.00 \][/tex]
### Comparison with [tex]\(K_p = 13.75\)[/tex]
Now we compare [tex]\(Q\)[/tex] to [tex]\(K_p\)[/tex] for each condition:
- Condition 1: [tex]\(Q = 112.50\)[/tex]; [tex]\(Q \neq K_p\)[/tex]
- Condition 2: [tex]\(Q = 16.00\)[/tex]; [tex]\(Q \neq K_p\)[/tex]
- Condition 3: [tex]\(Q = 16.67\)[/tex]; [tex]\(Q \neq K_p\)[/tex]
- Condition 4: [tex]\(Q = 100.00\)[/tex]; [tex]\(Q \neq K_p\)[/tex]
### Conclusion
Since [tex]\(Q\)[/tex] is not equal to [tex]\(K_p\)[/tex] (13.75) for any of the conditions, none of them are at equilibrium.
Thus, the conditions that are NOT at equilibrium are:
1. Condition 1
2. Condition 2
3. Condition 3
4. Condition 4
The reaction quotient [tex]\(Q\)[/tex] for this reaction is given by:
[tex]\[ Q = \frac{[HCl]^2}{[H_2] \cdot [Cl_2]} \][/tex]
Now, let's calculate [tex]\(Q\)[/tex] for each set of initial conditions and compare it with [tex]\(K_p = 13.75\)[/tex].
### Initial Conditions and Calculations
1. Condition 1:
- [tex]\( [H_2] = 0.1 \)[/tex]
- [tex]\( [Cl_2] = 0.2 \)[/tex]
- [tex]\( [HCl] = 1.5 \)[/tex]
[tex]\[ Q = \frac{(1.5)^2}{(0.1) \cdot (0.2)} = \frac{2.25}{0.02} = 112.50 \][/tex]
2. Condition 2:
- [tex]\( [H_2] = 0.5 \)[/tex]
- [tex]\( [Cl_2] = 0.5 \)[/tex]
- [tex]\( [HCl] = 2 \)[/tex]
[tex]\[ Q = \frac{(2)^2}{(0.5) \cdot (0.5)} = \frac{4}{0.25} = 16.00 \][/tex]
3. Condition 3:
- [tex]\( [H_2] = 0.3 \)[/tex]
- [tex]\( [Cl_2] = 0.2 \)[/tex]
- [tex]\( [HCl] = 1 \)[/tex]
[tex]\[ Q = \frac{(1)^2}{(0.3) \cdot (0.2)} = \frac{1}{0.06} = 16.67 \][/tex]
4. Condition 4:
- [tex]\( [H_2] = 0.4 \)[/tex]
- [tex]\( [Cl_2] = 0.4 \)[/tex]
- [tex]\( [HCl] = 4 \)[/tex]
[tex]\[ Q = \frac{(4)^2}{(0.4) \cdot (0.4)} = \frac{16}{0.16} = 100.00 \][/tex]
### Comparison with [tex]\(K_p = 13.75\)[/tex]
Now we compare [tex]\(Q\)[/tex] to [tex]\(K_p\)[/tex] for each condition:
- Condition 1: [tex]\(Q = 112.50\)[/tex]; [tex]\(Q \neq K_p\)[/tex]
- Condition 2: [tex]\(Q = 16.00\)[/tex]; [tex]\(Q \neq K_p\)[/tex]
- Condition 3: [tex]\(Q = 16.67\)[/tex]; [tex]\(Q \neq K_p\)[/tex]
- Condition 4: [tex]\(Q = 100.00\)[/tex]; [tex]\(Q \neq K_p\)[/tex]
### Conclusion
Since [tex]\(Q\)[/tex] is not equal to [tex]\(K_p\)[/tex] (13.75) for any of the conditions, none of them are at equilibrium.
Thus, the conditions that are NOT at equilibrium are:
1. Condition 1
2. Condition 2
3. Condition 3
4. Condition 4
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