Question

consider the following equilibrium:
2no(g) + cl₂(g) ⇌ 2nocl(g) δg⁰ = -41 kj
now suppose a reaction vessel is filled with 2.16 atm of chlorine (cl₂) and 0.272 atm of nitrosyl chloride (nocl) at 610. °c. answer the following questions about this system:
under these conditions, will the pressure of cl₂ 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 cl₂ will tend to rise, can that be changed to a tendency to fall by adding no? similarly, if you said the pressure of cl₂ 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:

Step1: Find \( K_p \) from \( \Delta G^0 \)

First, recall the relationship \( \Delta G^0 = -RT\ln K_p \). The temperature \( T = 610 + 273.15 = 883.15 \, \text{K} \), \( R = 8.314 \, \text{J/(mol·K)} = 0.008314 \, \text{kJ/(mol·K)} \), and \( \Delta G^0 = -41 \, \text{kJ} \).

Rearranging for \( \ln K_p \):
\( \ln K_p = -\frac{\Delta G^0}{RT} = -\frac{-41}{0.008314 \times 883.15} \approx \frac{41}{7.34} \approx 5.586 \)

Then \( K_p = e^{5.586} \approx 262 \).

Step2: Calculate reaction quotient \( Q_p \)

The reaction is \( 2\text{NO}(g) + \text{Cl}_2(g) ightleftharpoons 2\text{NOCl}(g) \). Initially, \( P_{\text{NO}} = 0 \) (since not given, assume initial NO pressure is 0? Wait, no—wait, the vessel is filled with \( \text{Cl}_2 = 2.16 \, \text{atm} \) and \( \text{NOCl} = 0.272 \, \text{atm} \), \( \text{NO} = 0 \) (initially, before reaction? Wait, no—wait, the initial pressures: \( P_{\text{NO}} = 0 \), \( P_{\text{Cl}_2} = 2.16 \, \text{atm} \), \( P_{\text{NOCl}} = 0.272 \, \text{atm} \).

The expression for \( Q_p \) is \( Q_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 (P_{\text{Cl}_2})} \). Since \( P_{\text{NO}} = 0 \), \( Q_p \) is undefined (infinite), but wait—no, maybe I misread. Wait, the problem says "a reaction vessel is filled with 2.16 atm of chlorine (\( \text{Cl}_2 \)) and 0.272 atm of nitrosyl chloride (\( \text{NOCl} \))"—so initial \( P_{\text{NO}} = 0 \), \( P_{\text{Cl}_2} = 2.16 \), \( P_{\text{NOCl}} = 0.272 \).

For the reaction, the direction: since \( Q_p \) (with \( P_{\text{NO}} = 0 \)) would imply the reaction proceeds to the left (because \( Q_p \) is infinite, much larger than \( K_p = 262 \)), so the reaction will shift left, producing \( \text{Cl}_2 \), so \( P_{\text{Cl}_2} \) will rise. Wait, but let's check again. Wait, if \( \text{NO} = 0 \), the reaction can't proceed right (needs NO), so it will proceed left: \( 2\text{NOCl} ightarrow 2\text{NO} + \text{Cl}_2 \), so \( \text{Cl}_2 \) pressure increases (rises).

Step3: Analyze effect of adding NO

Adding NO will increase the denominator of \( Q_p \), decreasing \( Q_p \). If we add enough NO, \( Q_p \) can become less than \( K_p \), so the reaction shifts right, consuming \( \text{Cl}_2 \), so \( P_{\text{Cl}_2} \) would fall. So yes, we can reverse the tendency.

Step4: Calculate minimum \( P_{\text{NO}} \) to reverse (make \( Q_p = K_p \))

We want \( Q_p = K_p \) (the point where the tendency reverses). Let \( P_{\text{NO}} = x \) (the minimum pressure needed). Then:

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

We know \( K_p = 262 \), \( P_{\text{NOCl}} = 0.272 \, \text{atm} \), \( P_{\text{Cl}_2} = 2.16 \, \text{atm} \). Rearranging for \( x \):

\( (x)^2 = \frac{(P_{\text{NOCl}})^2}{K_p \times P_{\text{Cl}_2}} \)

\( x^2 = \frac{(0.272)^2}{262 \times 2.16} \approx \frac{0.073984}{565.92} \approx 0.0001307 \)

\( x = \sqrt{0.0001307} \approx 0.01143 \, \text{atm} \)? Wait, that can't be right—wait, no, I messed up the direction. Wait, when adding NO, the reaction will shift right, so we need to consider the reaction quotient when the system is at equilibrium (or when \( Q_p = K_p \) for the reverse? Wait, no—wait, the initial reaction is shifting left (producing \( \text{Cl}_2 \)) because \( Q_p \) (with \( \text{NO} = 0 \)) is infinite. To make it shift right, we need \( Q_p < K_p \). Wait, no—wait, \( K_p \) is for the forward reaction. Let's re-express:

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

Initially, \( P_{\text{NO}} = 0 \), so \( Q_p \) is infinite (since denominator is 0), so reaction shifts left (reverse direction), producing \( \text{Cl}_2 \) (so \( P_{\text{Cl}_2} \) rises). To reverse this, we need the reaction to shift right, so \( Q_p < K_p \). For \( Q_p = K_p \), the system is at equilibrium. So we need to find \( P_{\text{NO}} \) such that \( Q_p = K_p \).

Wait, let's correct the \( Q_p \) calculation. Initially, \( P_{\text{NO}} = 0 \), \( P_{\text{Cl}_2} = 2.16 \), \( P_{\text{NOCl}} = 0.272 \). So \( Q_p = \frac{(0.272)^2}{(0)^2 (2.16)} \), which is undefined (infinite), so reaction proceeds left (since \( Q_p > K_p \), because \( K_p = 262 \), and infinite > 262). So to make \( Q_p < K_p \), we need to add NO (so \( P_{\text{NO}} > 0 \)), which increases the denominator, decreasing \( Q_p \).

Set \( Q_p = K_p \) (the threshold where the tendency reverses):

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

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

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

Plug in values:

\( (P_{\text{NO}})^2 = \frac{(0.272)^2}{262 \times 2.16} \)

Calculate numerator: \( 0.272^2 = 0.073984 \)

Denominator: \( 262 \times 2.16 = 565.92 \)

So \( (P_{\text{NO}})^2 = \frac{0.073984}{565.92} \approx 0.0001307 \)

\( P_{\text{NO}} = \sqrt{0.0001307} \approx 0.0114 \, \text{atm} \)? Wait, that seems too small. Wait, maybe I made a mistake in \( K_p \) calculation. Let's recheck \( \ln K_p \):

\( \Delta G^0 = -41 \, \text{kJ} \), \( R = 8.314 \, \text{J/(mol·K)} = 0.008314 \, \text{kJ/(mol·K)} \), \( T = 610 + 273 = 883 \, \text{K} \) (approx).

\( \ln K_p = -\Delta G^0/(RT) = -(-41)/(0.008314 \times 883) = 41/(7.34) \approx 5.586 \), so \( K_p = e^{5.586} \approx 262 \). That seems correct.

Wait, but if \( P_{\text{NO}} \) is 0, the reaction can't proceed forward, so it proceeds backward. When we add NO, the forward reaction becomes favorable. So the minimum \( P_{\text{NO}} \) is when \( Q_p = K_p \), so the reaction is at equilibrium (no net tendency). So solving for \( P_{\text{NO}} \):

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

Wait, but let's check units and logic again. The reaction is \( 2\text{NO} + \text{Cl}_2 ightleftharpoons 2\text{NOCl} \). So \( K_p = \frac{[NOCl]^2}{[NO]^2 [Cl_2]} \). So when \( [NO] \) increases, the denominator increases, so \( K_p > Q_p \), so reaction shifts right, consuming \( \text{Cl}_2 \), so \( P_{\text{Cl}_2} \) falls.

So with \( P_{\text{NOCl}} = 0.272 \), \( P_{\text{Cl}_2} = 2.16 \), \( K_p = 262 \):

\( [NO]^2 = \frac{[NOCl]^2}{K_p [Cl_2]} = \frac{(0.272)^2}{262 \times 2.16} \approx \frac{0.073984}{565.92} \approx 0.0001307 \)

\( [NO] = \sqrt{0.0001307} \approx 0.0114 \, \text{atm} \). Wait, that's very small, but mathematically it works. But let's check if this makes sense. If \( P_{\text{NO}} = 0.0114 \, \text{atm} \), then \( Q_p = \frac{(0.272)^2}{(0.0114)^2 (2.16)} \approx \frac{0.073984}{0.000130 \times 2.16} \approx \frac{0.073984}{0.0002808} \approx 263 \), which is approximately \( K_p = 262 \), so that's correct. So the minimum \( P_{\text{NO}} \) is approximately \( 0.011 \, \text{atm} \), but wait, let's recalculate:

\( (0.272)^2 = 0.073984 \)

\( 262 \times 2.16 = 565.92 \)

\( 0.073984 / 565.92 = 0.0001307 \)

Square root of 0.0001307: \( \sqrt{0.0001307} \approx 0.0114 \, \text{atm} \), which is \( 0.011 \, \text{atm} \) (two significant figures: 0.011 or 1.1×10⁻²? Wait, 0.0114 rounds to 0.011 (two sig figs) or 0.011 (since 1 and 1 are the sig figs). Wait, 0.0114 has three sig figs, so two sig figs is 0.011 or 1.1×10⁻².

But wait, maybe I made a mistake in the initial assumption. Wait, the problem says "a reaction vessel is filled with 2.16 atm of chlorine (Cl₂) and 0.272 atm of nitrosyl chloride (NOCl)". So initial NO pressure is 0? That seems odd, but maybe. So when NO is 0, the reaction proceeds backward (producing Cl₂), so Cl₂ pressure rises. Adding NO allows the forward reaction, so Cl₂ pressure falls. The minimum NO pressure is when Qp = Kp, so the reaction is at equilibrium (no net tendency). So the calculation gives ~0.011 atm.

Wait, but let's check the first question: "will the pressure of Cl₂ tend to rise or fall?" Since initial NO is 0, the reaction proceeds backward (2NOCl → 2NO + Cl₂), so Cl₂ pressure rises. So first answer: rise.

Second question: "Is it possible to reverse this tendency by adding NO?" Yes, because adding NO increases the forward reaction's reactant, shifting equilibrium right, consuming Cl₂, so Cl₂ pressure falls. So yes.

Third question: calculate minimum NO pressure. As above, ~0.011 atm (two sig figs: 0.011 or 1.1×10⁻²? Wait, 0.0114 rounds to 0.011 (two sig figs) or 0.011 (since the first non-zero digit is 1, then 1). Wait, 0.0114 is 1.14×10⁻², so two sig figs is 1.1×10⁻² or 0.011 atm.

But let's re-express the calculation:

\( P_{\text{NO}} = \sqrt{\frac{(P_{\text{NOCl}})^2}{K_p \cdot P_{\text{Cl}_2}}} = \sqrt{\frac{(0.272)^2}{262 \cdot 2.16}} \)

Calculate numerator: \( 0.272^2 = 0.073984 \)

Denominator: \( 262 \times 2.16 = 565.92 \)

Divide: \( 0.073984 / 565.92 = 0.0001307 \)

Square root: \( \sqrt{0.0001307} = 0.01143 \, \text{atm} \approx 0.011 \, \text{atm} \) (two sig figs).

Answers:

1. Under these conditions, the pressure of \( \text{Cl}_2 \) tends to **rise**.
2. Is it possible to reverse this tendency by adding NO? **Yes**.
3. The minimum pressure of NO needed to reverse it is \( \boldsymbol{0.011 \, \text{atm}} \) (or \( 1.1 \times 10^{-2} \, \text{atm} \), two significant figures).

Answer:

s:

1. Under these conditions, the pressure of \( \text{Cl}_2 \) tends to rise.
2. Is it possible to reverse this tendency by adding NO? Yes.
3. The minimum pressure of NO needed to reverse it is \( \boldsymbol{0.011 \, \text{atm}} \) (or \( 1.1 \times 10^{-2} \, \text{atm} \), two significant figures).