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Chapter 17: Problem 127

Consider the following reaction at \(35^{\circ} \mathrm{C} :\) $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \Delta G^{\circ}=20 . \mathrm{kJ} $$ If 2.0 atm of NOCl are reacted in a rigid container at $35^{\circ} \mathrm{C},$ calculate the equilibrium partial pressure of NO.

Short Answer

Expert verified
Therefore, the equilibrium partial pressure of NO is approximately 0.19 atm.

Step by step solution

01

Write the equilibrium constant expression in terms of pressures

First, we need to write the expression for the equilibrium constant (Kp) in terms of partial pressures. For the given reaction: \(2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Cl}_{2}(g)\) The expression for Kp is: \(K_{p} = \frac{P_{\mathrm{NO}}^2 \times P_{\mathrm{Cl}_{2}}}{P_{\mathrm{NOCl}}^2}\)
02

Calculate Kp from ΔG°

Next, we need to calculate Kp using the given standard Gibbs free energy change (ΔG°) of the reaction. We can use the following equation: \(\Delta G^{\circ} = -RT \ln{K_{p}}\) For the given problem, ΔG° = 20 kJ, which is equal to 20,000 J. Also, T = 35°C, which is equal to 308 K (converting from Celsius to Kelvin). We know that the universal gas constant, R = 8.314 J/(mol·K). Now, we can rearrange the equation to solve for Kp: \(K_{p} = e^{(-\Delta G^{\circ}/RT)} = e^{(-20,000/(8.314 \times 308))}\)
03

Calculate the equilibrium partial pressures using ICE table

To find the equilibrium partial pressures of the reactants and products, we can use an ICE (Initial, Change, Equilibrium) table. At the beginning of the reaction, the partial pressures are: NOCl = 2.0 atm, NO = 0 atm, Cl₂ = 0 atm Since the reaction has a 2:2:1 stoichiometry, during the reaction, the change in partial pressures will be: NOCl = -2x, NO = +2x, and Cl₂ = +x At equilibrium, the partial pressures are: NOCl = 2.0-2x, NO = 2x, and Cl₂ = x Now, we can plug these values into the Kp expression: \(K_{p} = \frac{(2x)^2 \times x}{(2.0-2x)^2}\)
04

Solve for x and find the equilibrium partial pressure of NO

Substitute the value of Kp obtained in Step 2 into the above equation: \(K_{p} = \frac{(2x)^2 \times x}{(2.0-2x)^2}\) Now, solve for x using numerical methods or a suitable solver: x ≈ 0.095 With the value of x, we can calculate the equilibrium partial pressure of NO: P_NO = 2x ≈ 2 * 0.095 ≈ 0.19 atm Therefore, the equilibrium partial pressure of NO is approximately 0.19 atm.

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Most popular questions from this chapter

Given the values of \(\Delta H\) and \(\Delta S,\) which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. $\Delta H=+25 \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}$ b. $\Delta H=+25 \mathrm{kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}$ c. $\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=298 \mathrm{K}$ d. $\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}$

Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and water at \(0^{\circ} \mathrm{C}\) . Vessel 2 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and a saltwater solution at \(0^{\circ} \mathrm{C}\) . Consider the process $\mathrm{H}_{2} \mathrm{O}(s) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)$ a. Determine the sign of \(\Delta S, \Delta S_{\text { sum }}\) and $\Delta S_{\text { univ }}$ for the process in vessel 1 . b. Determine the sign of \(\Delta S, \Delta S_{\text { sum }},\) and $\Delta S_{\text { univ }}\( for the process in vessel \)2 .$ (Hint: Think about the effect that a salt has on the freezing point of a solvent.)

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C},\) the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

For rubidium \(\Delta H_{\mathrm{vap}}^{\circ}=69.0 \mathrm{kJ} / \mathrm{mol}\) \(686^{\circ} \mathrm{C},\) its boiling point. Calculate $\Delta S^{\circ}, q, w,\( and \)\Delta E$ for the vaporization of 1.00 mole of rubidium at \(686^{\circ} \mathrm{C}\) and 1.00 atm pressure.

Consider the following reaction at 298 K: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ An equilibrium mixture contains \(\mathrm{O}_{2}(g)\) and \(\mathrm{SO}_{3}(g)\) at partial pressures of 0.50 \(\mathrm{atm}\) and 2.0 \(\mathrm{atm}\) , respectively. Using data from Appendix \(4,\) determine the equilibrium partial pressure of \(\mathrm{SO}_{2}\) in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?
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