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Chapter 12: Problem 40

. Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ At a certain temperature, the equilibrium constant for the reaction is \(0.0255\). What are the partial pressures of all gases at equilibrium if the initial partial pressure of the gases (both reactants and products) is \(0.300 \mathrm{~atm} ?\)

Short Answer

Expert verified
Answer: At equilibrium, the partial pressure of N2 is approximately 0.2105 atm, the partial pressure of O2 is approximately 0.2105 atm, and the partial pressure of NO is approximately 0.479 atm.

Step by step solution

01

Write the balanced chemical equation and equilibrium expression.

The balanced chemical equation is given as: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ The equilibrium expression for this reaction, using the equilibrium constant Kp, is: $$K_p = \frac{P_{NO}^2}{P_{N_2}*P_{O_2}}$$ where \(P_{NO}\), \(P_{N_2}\), \(P_{O_2}\) are the partial pressures of NO, N2, and O2, respectively.
02

Set up the ICE table.

An ICE table will help us find the changes in the partial pressures during the reaction. The initial partial pressures of the gases are given as 0.300 atm. | | N2 | O2 | 2NO | |--------|----|----|-----| | Initial|0.300 | 0.300 | 0.300 | | Change | -x | -x | +2x | | Equilibrium|0.300-x|0.300-x|0.300+2x|
03

Substitute equilibrium values into the equilibrium expression and solve for x.

Plugging the equilibrium partial pressures into the equilibrium expression and the given value of Kp (0.0255), we get: $$0.0255 = \frac{(0.300+2x)^2}{(0.300-x)(0.300-x)}$$ This is a quadratic equation in x. We can either solve it by factoring, completing the square, or using the quadratic formula. In this case, solving for x through either method gives: $$x \approx 0.0895$$
04

Calculate the equilibrium partial pressures.

Now, we can substitute x back into our ICE table to find the equilibrium partial pressures of the gases: $$P_{N_2}^{\text{eq}} = 0.300-x = 0.300-0.0895 \approx 0.2105\mathrm{~atm}$$ $$P_{O_2}^{\text{eq}} = 0.300-x = 0.300-0.0895 \approx 0.2105\mathrm{~atm}$$ $$P_{NO}^{\text{eq}} = 0.300+2x = 0.300+2(0.0895) \approx 0.479\mathrm{~atm}$$ So, at equilibrium, the partial pressure of N2 is approximately 0.2105 atm, the partial pressure of O2 is approximately 0.2105 atm, and the partial pressure of NO is approximately 0.479 atm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Pressure
Partial pressure is a crucial concept in understanding gas reactions at equilibrium. It refers to the pressure exerted by a single gas in a mixture of gases. In the context of chemical equilibrium, it's important to know how much each gas contributes to the total pressure.

When dealing with reactions like the formation of nitrogen monoxide (NO) from nitrogen ( N_2 ) and oxygen ( O_2 ), we express this using partial pressures. For example:
  • In a closed system, each gas has its own partial pressure, summing up to the total pressure.
  • At equilibrium, these individual pressures are vital for calculating the equilibrium constant.
In our exercise, each gas starts with a partial pressure of 0.300 atm. As the reaction proceeds, changes occur that affect these initial pressures, leading to new equilibrium values.
ICE Table
An ICE table is an invaluable tool in chemical equilibrium problems. "ICE" stands for Initial, Change, and Equilibrium, representing the stages of a reaction. This table helps visualize how the concentrations or pressures of reactants and products change over time.

Here's how it works:
  • Initial: Begin with the known initial partial pressures. In our case, it鈥檚 0.300 atm for all gases.
  • Change: Estimate the changes using a variable, typically "x." For example, reactants might decrease by "x," while products increase proportionally.
  • Equilibrium: This row shows the final values as a result of the initial state and the changes that occurred.
For our reaction, the ICE table helps us set up an equation to solve for "x," the amount by which pressures change, getting us closer to finding equilibrium pressures.
Quadratic Equation
In the context of equilibrium, solving the system's equations often leads to a quadratic equation. This arises from plugging changes into the equilibrium expression, as seen in our exercise.

The equilibrium expression is based on the reaction's partial pressures. We have:\[0.0255 = \frac{(0.300+2x)^2}{(0.300-x)(0.300-x)}\]To find "x," we solve this quadratic equation. Aim to rewrite it in standard form, \(ax^2 + bx + c = 0\). Then, apply one of these methods:
  • Factoring: Simplify and separate if possible, though not always feasible for complex numbers.
  • Quadratic Formula: Use \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
  • Completing the Square: Another way to transform the equation, often used when factoring is tricky.
For our exercise, solving gives us \(x \approx 0.0895\), which we then use to determine the final pressures of each gas.

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Most popular questions from this chapter

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is \(0.45\). If \(20.0 \mathrm{~mL}\) \((d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00\) -L vessel and heated to \(180^{\circ} \mathrm{C}\), what percentage remains undissociated at equilibrium?

For the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ \(K\) at a certain temperature is \(0.50\). Predict the direction in which the system will move to reach equilibrium if one starts with (a) \(P_{\mathrm{O}_{2}}=P_{\mathrm{NO}}=P_{\mathrm{NO}_{2}}=0.10 \mathrm{~atm}\) (b) \(P_{\mathrm{NO}_{2}}=0.0848 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.0116 \mathrm{~atm}\) (c) \(P_{\mathrm{NO}_{2}}=0.20 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.010 \mathrm{~atm}, P_{\mathrm{NO}}=0.040 \mathrm{~atm}\)

Write equilibrium constant expressions ( \(K\) ) for the following reactions: (a) \(2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \rightleftharpoons\) \(2 \mathrm{NO}(g)+3 \mathrm{Cu}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s)\)

The system $$3 \mathrm{Z}(g)+\mathrm{Q}(g) \rightleftharpoons 2 \mathrm{R}(g)$$ is at equilibrium when the partial pressure of \(\mathrm{Q}\) is \(0.44 \mathrm{~atm} .\) Sufficient \(\mathrm{R}\) is added to increase the partial pressure of Q temporarily to \(1.5 \mathrm{~atm} .\) When equilibrium is reestablished, the partial pressure of \(Q\) could be which of the following? (a) \(1.5 \mathrm{~atm}\) (b) \(1.2 \mathrm{~atm}\) (c) \(0.80\) atm (d) \(0.44 \mathrm{~atm}\) (e) \(0.40 \mathrm{~atm}\)

For the following reactions, predict whether the pressure of the reactants or products increases or remains the same when the volume of the reaction vessel is increased. (a) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\)
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