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Chapter 12: Problem 51

At a particular temperature, \(K=3.75\) for the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)$$.If all four gases had initial concentrations of 0.800 \(M,\) calculate the equilibrium concentrations of the gases.

Short Answer

Expert verified
The equilibrium concentrations of the gases involved in the reaction are: \([\mathrm{SO}_2]_{eq} = [\mathrm{NO}_{2}]_{eq} = 0.185\ M\) and \([\mathrm{SO}_3]_{eq} = [\mathrm{NO}]_{eq} = 1.415\ M\).

Step by step solution

01

Write down the balanced chemical equation

We are already given a balanced chemical equation: \(\mathrm{SO}_{2}(g) + \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) + \mathrm{NO}(g)\)
02

Write the K expression for the reaction

Based on the reaction, the K expression can be written as: \(K = \frac{[\mathrm{SO}_3][\mathrm{NO}]}{[\mathrm{SO}_2][\mathrm{NO}_2]}\) where [A] represents the concentration of any species A in the mixture.
03

Set up the ICE table

We will set up an ICE table to represent the initial concentrations, changes, and equilibrium concentrations for each gas involved in the reaction. ``` | SO₂ | NO₂ | SO₃ | NO Initial | 0.8 | 0.8 | 0.8 | 0.8 Change | -x | -x | +x | +x Equilibrium | 0.8-x | 0.8-x | 0.8+x | 0.8+x ```
04

Substitute the ICE table values into the K expression

Now we will substitute the equilibrium concentrations from the ICE table into the K expression: \(3.75 = \frac{(0.8+x)(0.8+x)}{(0.8-x)(0.8-x)}\)
05

Solve the equation for x

This is a quadratic equation in terms of x. To solve this equation, we can try multiplying both sides by the denominator and simplifying the equation: \(3.75(0.8 - x)^2 = (0.8 + x)^2\) Expand and simplify the equation to get: \(3 = 5x^2 - 8x\) Rearrange the equation as a standard quadratic equation: \(5x^2 - 8x - 3 = 0\) Now, solve the equation for x using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Plugging in the values, we get: \(x = \frac{8 \pm \sqrt{(-8)^2 - 4(5)(-3)}}{2(5)}\) Solving for x, we get two possible values: \(x = -0.215\) and \(x = 0.615\). Since we can't have a negative change in concentration, we will use the positive value for x: \(x = 0.615\)
06

Calculate the equilibrium concentrations

Finally, use the value of x to calculate the equilibrium concentrations for each gas: - \(\mathrm{SO}_2\) and \(\mathrm{NO}_2\): \(0.8 - 0.615 = 0.185\ mol\,L^{-1}\) - \(\mathrm{SO}_3\) and \(\mathrm{NO}\): \(0.8 + 0.615 = 1.415\ mol\,L^{-1}\) So, the equilibrium concentrations of the gases are: - \([\mathrm{SO}_2]_{eq} = [\mathrm{NO}_{2}]_{eq} = 0.185\ M\) - \([\mathrm{SO}_3]_{eq} = [\mathrm{NO}]_{eq} = 1.415\ M\)

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Most popular questions from this chapter

For the reaction:$$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$. \(K=1.8 \times 10^{-7}\) at a certain temperature. If at equilibrium \(\left[\mathrm{O}_{2}\right]=0.062 \mathrm{M},\) calculate the equilibrium \(\mathrm{O}_{3}\) concentration.

Consider the reaction $$\mathbf{P}_{4}(g) \longrightarrow 2 \mathbf{P}_{2}(g)$$, where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at \(1325 \mathrm{K}\). In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at \(1325 \mathrm{K},\) the equilibrium mixture of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g)\) has a total pressure of 1.00 atm. Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(P_{4}(g)\) that has dissociated to reach equilibrium.

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}} \approx 1 \times 10^{-31}\) for the reaction $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$. a. Calculate the concentration of NO, in molecules/cm \(^{3}\) that can exist in equilibrium in air at \(25^{\circ} \mathrm{C} .\) In air, \(P_{\mathrm{N}_{2}}=0.8 \mathrm{atm}\) and \(P_{\mathrm{O}_{2}}=0.2 \mathrm{atm}\). b. Typical concentrations of NO in relatively pristine environments range from \(10^{8}\) to \(10^{10}\) molecules/cm \(^{3}\). Why is there a discrepancy between these values and your answer to part a?

In a solution with carbon tetrachloride as the solvent, the compound VCl_ undergoes dimerization:$$2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2} \mathrm{Cl}_{8}$$ When \(6.6834 \mathrm{g} \mathrm{VCl}_{4}\) is dissolved in \(100.0 \mathrm{g}\) carbon tetrachloride, the freezing point is lowered by \(5.97^{\circ} \mathrm{C}\). Calculate the value of the equilibrium constant for the dimerization of \(\mathrm{VCl}_{4}\) at this temperature. (The density of the equilibrium mixture is \(\left.1.696 \mathrm{g} / \mathrm{cm}^{3}, \text { and } K_{\mathrm{f}}=29.8^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol} \text { for } \mathrm{CCl}_{4} .\right)\).

Hydrogen for use in ammonia production is produced by the reaction $$ \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\text { Ni catalyst }}{=\frac{750^{\circ} \mathrm{C}}{7}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) $$ What will happen to a reaction mixture at equilibrium if a. \(\mathrm{H}_{2} \mathrm{O}(g)\) is removed? b. the temperature is increased (the reaction is endothermic)? c. an inert gas is added to a rigid reaction container? d. \(\mathrm{CO}(g)\) is removed? e. the volume of the container is tripled?
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