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Chapter 12: Problem 53

At \(2200^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.050\) for the reaction $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$.What is the partial pressure of NO in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that were placed in a flask at initial pressures of 0.80 and 0.20 atm, respectively?

Short Answer

Expert verified
The partial pressure of NO in equilibrium with N2 and O2, given their initial pressures of 0.80 atm and 0.20 atm, respectively, at a temperature of 2200°C and \(K_p = 0.050\), is approximately 0.028 atm.

Step by step solution

01

Identify the Balanced Equation and Equilibrium Constant

We are given the balanced equation: \[\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}(g)\] At 2200°C, the equilibrium constant \(K_p\) is given as 0.050.
02

Define the Initial and Change in Pressures

The initial pressures of N2 and O2 are given as 0.80 atm and 0.20 atm, respectively. Initially, there is no NO present. Let's denote the change in pressure for each component as follows: - N2: -x (pressure decreases as it reacts) - O2: -x (pressure decreases as it reacts) - NO: +2x (pressure increases as it forms)
03

Write Equilibrium Pressures

Using the expressions in Step 2, we can write the equilibrium pressures for each component: - N2: 0.80 - x - O2: 0.20 - x - NO: 2x
04

Set Up the Equilibrium Constant Expression

Now, let's set up the equilibrium constant expression using the equilibrium constant value and the pressures for each component: \[K_p = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_2][\mathrm{O}_2]}\] Substitute the values of the equilibrium pressures: \[0.050 = \frac{(2x)^2}{(0.80 - x)(0.20 - x)}\]
05

Solve for x

Solve the above equation for the value of x: \[0.050 = \frac{4x^2}{(0.80 - x)(0.20 - x)}\] It's a quadratic equation, and solving it using any preferred method (such as factoring, quadratic formula, or numerically) gives: x ≈ 0.014
06

Determine the Equilibrium Pressure of NO

With the value of x, we can now find the equilibrium pressure for NO: Partial pressure of NO = 2x = 2 × 0.014 ≈ 0.028 atm Therefore, the partial pressure of NO in equilibrium with N2 and O2 is approximately 0.028 atm.

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Most popular questions from this chapter

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\P_{\mathrm{NH}}=3.1 \times 10^{-2} \mathrm{atm} \\\P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm} \\\P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}\end{array}$$.Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525\) atm, \(P_{\mathrm{NH},}=0.0167\) atm, and \(P_{\mathrm{H}_{2}}=0.00761\) atm, does this represent a system at equilibrium?

An equilibrium mixture contains 0.60 g solid carbon and the gases carbon dioxide and carbon monoxide at partial pressures of 2.60 atm and 2.89 atm, respectively. Calculate the value of \(K_{\mathrm{p}}\) for the reaction \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\).

An \(8.00-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) was placed in an evacuated container, where it decomposed at \(600^{\circ} \mathrm{C}\) according to the following reaction: $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ At equilibrium the total pressure and the density of the gaseous mixtures were 1.80 atm and \(1.60 \mathrm{g} / \mathrm{L},\) respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.

For the reaction $$2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$$,\(K=2.4 \times 10^{-3}\) at a given temperature. At equilibrium in a \(2.0-\mathrm{L}\) container it is found that \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=1.1 \times 10^{-1} \mathrm{M}\) and \(\left[\mathrm{H}_{2}(g)\right]=1.9 \times 10^{-2} \mathrm{M} .\) Calculate the moles of \(\mathrm{O}_{2}(g)\) present under these conditions.

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\).Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\). b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\). d. \(2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\).
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