Question

consider the following equilibrium:
2nocl(g) ⇌ 2no(g) + cl₂(g) δg⁰ = 41 kj
now suppose a reaction vessel is filled with 1.47 atm of nitrosyl chloride (nocl) and 6.16 atm of chlorine (cl₂) at 892. °c. answer the following questions about this system:
under these conditions, will the pressure of nocl tend to rise or fall? ∘ rise ∘ fall
is it possible to reverse this tendency by adding no?
in other words, if you said the pressure of nocl will tend to rise, can that be changed to a tendency to fall by adding no? similarly, if you said the pressure of nocl will tend to fall, can that be changed to a tendency to rise by adding no? ∘ yes ∘ no
if you said the tendency can be reversed in the second question, calculate the minimum pressure of no needed to reverse it.
round your answer to 2 significant digits atm

Explanation:

Step 1: Find \( K_p \) from \( \Delta G^0 \)

First, recall the relationship \( \Delta G^0 = -RT\ln K_p \). The temperature \( T = 892 + 273.15 = 1165.15 \, \text{K} \), \( R = 8.314 \, \text{J/(mol·K)} = 0.008314 \, \text{kJ/(mol·K)} \), and \( \Delta G^0 = 41 \, \text{kJ/mol} \).

Rearranging for \( \ln K_p \):
\( \ln K_p = \frac{-\Delta G^0}{RT} = \frac{-41}{0.008314 \times 1165.15} \approx \frac{-41}{9.69} \approx -4.23 \)

Then \( K_p = e^{-4.23} \approx 0.0159 \).

Step 2: Calculate reaction quotient \( Q_p \)

The reaction is \( 2\text{NOCl}(g) ightleftharpoons 2\text{NO}(g) + \text{Cl}_2(g) \). Initially, \( P_{\text{NOCl}} = 1.47 \, \text{atm} \), \( P_{\text{Cl}_2} = 6.16 \, \text{atm} \), \( P_{\text{NO}} = 0 \, \text{atm} \).

\( Q_p = \frac{(P_{\text{NO}})^2 P_{\text{Cl}_2}}{(P_{\text{NOCl}})^2} = \frac{(0)^2 \times 6.16}{(1.47)^2} = 0 \).

Step 3: Compare \( Q_p \) and \( K_p \)

Since \( Q_p (0) < K_p (0.0159) \), the reaction will proceed forward (toward products). So \( \text{NOCl} \) is consumed? Wait, no—wait, forward reaction: \( \text{NOCl} \) decomposes into \( \text{NO} \) and \( \text{Cl}_2 \). Wait, initial \( Q_p = 0 \), so reaction goes forward (products form). So \( \text{NOCl} \) is being consumed? Wait, no: the reaction is \( 2\text{NOCl} ightarrow 2\text{NO} + \text{Cl}_2 \). So if \( Q_p < K_p \), reaction proceeds right (forward), so \( \text{NOCl} \) pressure falls? Wait, no—wait, when reaction proceeds forward, \( \text{NOCl} \) is consumed, so its pressure falls? Wait, but initial \( P_{\text{NO}} = 0 \), so \( Q_p = 0 \), which is less than \( K_p \), so reaction moves forward (toward products), so \( \text{NOCl} \) is converted to \( \text{NO} \) and \( \text{Cl}_2 \), so \( P_{\text{NOCl}} \) falls? Wait, but the first question: "will the pressure of \( \text{NOCl} \) tend to rise or fall?" Wait, maybe I messed up. Wait, \( Q_p = \frac{[NO]^2 [Cl_2]}{[NOCl]^2} \). Initially, \( [NO] = 0 \), so \( Q_p = 0 \). \( K_p = 0.0159 \), so \( Q_p < K_p \), so reaction shifts right (forward), so \( \text{NOCl} \) is consumed (decomposed), so its pressure falls? Wait, but that would mean \( P_{\text{NOCl}} \) falls. Wait, but let's check again.

Wait, maybe I made a mistake in \( K_p \) sign. Wait, \( \Delta G^0 = 41 \, \text{kJ} \) (positive), so \( K_p < 1 \), which matches \( K_p \approx 0.0159 \). So \( Q_p = 0 < K_p \), so reaction proceeds forward (toward products), so \( \text{NOCl} \) is consumed, so its pressure falls? Wait, but the first question: "will the pressure of \( \text{NOCl} \) tend to rise or fall?" So if reaction goes forward, \( \text{NOCl} \) is used up, so pressure falls? Wait, no—wait, \( 2\text{NOCl} \) produces \( 2\text{NO} + \text{Cl}_2 \). So when \( \text{NOCl} \) decomposes, its pressure decreases. So the answer to the first question is "fall".

Step 4: Effect of adding \( \text{NO} \)

Adding \( \text{NO} \) increases \( P_{\text{NO}} \), which increases \( Q_p \) (since \( Q_p \propto [NO]^2 \)). If we add enough \( \text{NO} \), \( Q_p \) can become greater than \( K_p \), causing the reaction to shift left (toward \( \text{NOCl} \)), which would increase \( P_{\text{NOCl}} \) (reverse the tendency). So the answer to the second question is "yes".

Step 5: Calculate minimum \( P_{\text{NO}} \) to reverse the tendency

We want the reaction to shift left, so \( Q_p \geq K_p \). Let \( P_{\text{NO}} = x \). Then:

\( Q_p = \frac{(x)^2 \times 6.16}{(1.47)^2} \geq K_p = 0.0159 \)

Solve for \( x \):

\( x^2 \geq \frac{0.0159 \times (1.47)^2}{6.16} \)

Calculate denominator: \( (1.47)^2 = 2.1609 \), numerator: \( 0.0159 \times 2.1609 \approx 0.0344 \), then \( \frac{0.0344}{6.16} \approx 0.00558 \)

So \( x^2 \geq 0.00558 \), \( x \geq \sqrt{0.00558} \approx 0.0747 \, \text{atm} \)? Wait, no—wait, to reverse the tendency, we need \( Q_p > K_p \) (so reaction shifts left, increasing \( P_{\text{NOCl}} \)). Wait, initial tendency is \( Q_p < K_p \) (shift right, decreasing \( P_{\text{NOCl}} \)). To reverse it, we need \( Q_p > K_p \) (shift left, increasing \( P_{\text{NOCl}} \)).

So set \( Q_p = K_p \) (minimum \( P_{\text{NO}} \) to make \( Q_p \geq K_p \)):

\( \frac{(x)^2 \times 6.16}{(1.47)^2} = 0.0159 \)

\( x^2 = \frac{0.0159 \times (1.47)^2}{6.16} \)

Calculate:

\( (1.47)^2 = 2.1609 \)

\( 0.0159 \times 2.1609 \approx 0.0344 \)

\( \frac{0.0344}{6.16} \approx 0.00558 \)

\( x = \sqrt{0.00558} \approx 0.0747 \, \text{atm} \)? Wait, that seems too low. Wait, maybe I messed up the direction. Wait, initial \( Q_p = 0 \) (since \( P_{\text{NO}} = 0 \)). To reverse the tendency (from falling to rising), we need the reaction to shift left, so \( Q_p > K_p \). Let's re-express:

Reaction: \( 2\text{NOCl} ightleftharpoons 2\text{NO} + \text{Cl}_2 \)

\( K_p = \frac{(P_{\text{NO}})^2 P_{\text{Cl}_2}}{(P_{\text{NOCl}})^2} = 0.0159 \)

We want the reaction to shift left, so \( Q_p > K_p \). Let \( P_{\text{NO}} = x \), \( P_{\text{NOCl}} = 1.47 \) (initial, but when we add \( \text{NO} \), the reaction will shift, but we can assume initial \( P_{\text{NOCl}} = 1.47 \), \( P_{\text{Cl}_2} = 6.16 \), and \( P_{\text{NO}} = x \) (added)).

So \( Q_p = \frac{x^2 \times 6.16}{(1.47)^2} \). Set \( Q_p = K_p \) (minimum \( x \) to make \( Q_p \geq K_p \)):

\( x^2 = \frac{K_p \times (1.47)^2}{6.16} = \frac{0.0159 \times 2.1609}{6.16} \approx \frac{0.0344}{6.16} \approx 0.00558 \)

\( x = \sqrt{0.00558} \approx 0.0747 \, \text{atm} \)? Wait, that seems too small. Wait, maybe I made a mistake in \( K_p \) calculation. Let's recalculate \( \ln K_p \):

\( \Delta G^0 = 41000 \, \text{J/mol} \), \( R = 8.314 \, \text{J/(mol·K)} \), \( T = 892 + 273 = 1165 \, \text{K} \)

\( \ln K_p = \frac{-41000}{8.314 \times 1165} = \frac{-41000}{9686} \approx -4.23 \), so \( K_p = e^{-4.23} \approx 0.0159 \). That's correct.

Wait, maybe the initial tendency: when \( Q_p = 0 \), reaction shifts right (produces \( \text{NO} \) and \( \text{Cl}_2 \)), so \( P_{\text{NOCl}} \) falls (since it's consumed). So the first answer is "fall".

To reverse the tendency (make \( P_{\text{NOCl}} \) rise), we need the reaction to shift left, which requires \( Q_p > K_p \). Adding \( \text{NO} \) increases \( Q_p \), so yes, it's possible.

Now, calculate the minimum \( P_{\text{NO}} \) such that \( Q_p = K_p \) (the point where the tendency reverses). So:

\( \frac{(P_{\text{NO}})^2 \times 6.16}{(1.47)^2} = 0.0159 \)

Solve for \( P_{\text{NO}} \):

\( (P_{\text{NO}})^2 = \frac{0.0159 \times (1.47)^2}{6.16} \)

\( (1.47)^2 = 2.1609 \)

\( 0.0159 \times 2.1609 = 0.0344 \)

\( \frac{0.0344}{6.16} = 0.00558 \)

\( P_{\text{NO}} = \sqrt{0.00558} \approx 0.0747 \, \text{atm} \approx 0.075 \, \text{atm} \) (2 significant digits). Wait, but this seems very low. Did I mix up the reaction direction?

Wait, maybe the reaction is \( 2\text{NO} + \text{Cl}_2 ightleftharpoons 2\text{NOCl} \), so \( K_p' = 1/K_p \approx 62.9 \). Let's check: if \( \Delta G^0 = 41 \, \text{kJ} \) for the forward reaction (decomposition), then for the reverse reaction (formation), \( \Delta G^0 = -41 \, \text{kJ} \), so \( K_p' = e^{41000/(8.314 \times 1165)} = e^{4.23} \approx 68.6 \). Ah! Here's the mistake: I used the wrong reaction direction for \( K_p \). The given reaction is \( 2\text{NOCl} ightleftharpoons 2\text{NO} + \text{Cl}_2 \), so \( K_p \) is for this reaction. But when we calculate \( Q_p \), we should use the same reaction. Wait, but if \( \Delta G^0 \) is positive, \( K_p < 1 \), so the forward reaction is non-spontaneous, and the reverse reaction is spontaneous. Wait, initial \( Q_p = 0 \), so the reaction will proceed forward (toward products) because \( Q_p < K_p \), even though \( \Delta G^0 \) is positive. Wait, \( \Delta G = \Delta G^0 + RT\ln Q_p \). At \( Q_p = 0 \), \( \ln Q_p = -\infty \), so \( \Delta G = -\infty \), which means the forward reaction is spontaneous (even though \( \Delta G^0 \) is positive). So \( \text{NOCl} \) will decompose, so its pressure falls (answer: fall).

Now, adding \( \text{NO} \) increases \( Q_p \). We want to make \( \Delta G > 0 \) (reaction spontaneous in reverse, so \( \text{NOCl} \) pressure rises). So \( \Delta G = \Delta G^0 + RT\ln Q_p > 0 \).

\( \ln Q_p > \frac{-\Delta G^0}{RT} = -4.23 \) (from before), so \( Q_p > e^{-4.23} \approx 0.0159 \).

\( Q_p = \frac{(P_{\text{NO}})^2 P_{\text{Cl}_2}}{(P_{\text{NOCl}})^2} \). Let \( P_{\text{NO}} = x \), \( P_{\text{NOCl}} = 1.47 \), \( P_{\text{Cl}_2} = 6.16 \).

So \( \frac{x^2 \times 6.16}{(1.47)^2} > 0.0159 \)

Solve for \( x \):

\( x^2 > \frac{0.0159 \times (1.47)^2}{6.16} \approx 0.00558 \)

\( x > \sqrt{0.00558} \approx 0.0747 \, \text{atm} \). So the minimum \( P_{\text{NO}} \) is approximately \( 0.075 \, \text{atm} \) (2 significant digits).

Answers:

1. Under these conditions, the pressure of NOCl tends to **fall**.
2. Is it possible to reverse this tendency by adding NO? **Yes**.
3. Minimum \( P_{\text{NO}} \) needed: \( \approx 0.075 \, \text{atm} \) (or 0.074–0.075 atm, rounded to 2 significant digits: 0.075 or 0.074, but let's recalculate precisely):

\( \frac{0.0159 \times (1.47)^2}{6.16} = \frac{0.0159 \times 2.1609}{6.16} = \frac{0.03436}{6.16} \approx 0.005578 \)

\( \sqrt{0.005578} \approx 0.0747 \approx 0.075 \, \text{atm} \).

Final Answers:

1. fall
2. yes
3. \( \boxed{0.075} \) (or 0.074–0.075, depending on rounding)

Answer:

s:

1. Under these conditions, the pressure of NOCl tends to fall.
2. Is it possible to reverse this tendency by adding NO? Yes.
3. Minimum \( P_{\text{NO}} \) needed: \( \approx 0.075 \, \text{atm} \) (or 0.074–0.075 atm, rounded to 2 significant digits: 0.075 or 0.074, but let's recalculate precisely):

\( \frac{0.0159 \times (1.47)^2}{6.16} = \frac{0.0159 \times 2.1609}{6.16} = \frac{0.03436}{6.16} \approx 0.005578 \)

\( \sqrt{0.005578} \approx 0.0747 \approx 0.075 \, \text{atm} \).

Final Answers:

1. fall
2. yes
3. \( \boxed{0.075} \) (or 0.074–0.075, depending on rounding)