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Chapter 14: Problem 108

For the reaction \(2 \mathrm{NO}_{2}(g) \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{4}(g)\) \(K_{\text {eq }}=0.500\). What is the equilibrium molar concentration of \(\mathrm{NO}_{2}(g)\) if \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.248 \mathrm{M}\) ?

Short Answer

Expert verified
The equilibrium molar concentration of NO₂(g) is approximately 0.704 M.

Step by step solution

01

Write the equilibrium constant expression for the reaction

The given reaction is: \[2 \mathrm{NO}_{2}(g) \rightleftarrows \mathrm{N}_{2}\mathrm{O}_{4}(g)\] First, we need to write the equilibrium constant expression for this reaction. Since it's a gaseous reaction, we use the molar concentrations (represented by square brackets) to denote activity: \[K_{eq} = \frac{[\mathrm{N}_{2}\mathrm{O}_{4}]}{[\mathrm{NO}_{2}]^2}\]
02

Substitute the given values

We are given that \([N_2O_4]=0.248 M\) and \(K_{eq}=0.500\). We can substitute these values into the equilibrium constant expression: \[0.500 = \frac{0.248}{[\mathrm{NO}_{2}]^2}\]
03

Solve for the molar concentration of NO₂

We need to solve for \([\mathrm{NO}_{2}]\) in the above equation. To do that, follow these steps: 1. Multiply both sides of the equation by \([\mathrm{NO}_{2}]^2\): \[[\mathrm{NO}_{2}]^2(0.500) = 0.248\] 2. Divide both sides of the equation by \(0.500\): \[[\mathrm{NO}_{2}]^2 = \frac{0.248}{0.500}\] 3. Calculate the division result: \[[\mathrm{NO}_{2}]^2 = 0.496\] 4. Take the square root of both sides of the equation to find the value of \([\mathrm{NO}_{2}]\): \[[\mathrm{NO}_{2}] = \sqrt{0.496}\] \[[\mathrm{NO}_{2}] \approx 0.704\] So, the equilibrium molar concentration of NO₂(g) is approximately 0.704 M.

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Most popular questions from this chapter

Consider the reaction \(\mathrm{SnO}_{2}(s)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{Sn}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) run in an explosion-proof sealed vessel. (a) Running the reaction in a sealed vessel allows equilibrium to be established. Explain why. (b) Express the concentration of \(\mathrm{H}_{2}(g)\) in terms of \(K_{\text {eq }}\) (c) The equilibrium constant for this reaction decreases as the reaction mixture is heated. Which way does the equilibrium shift? (d) Is the reaction exothermic or endothermic? Explain. (e) A new employee at a chemical company adds additional \(\mathrm{SnO}_{2}(s)\) to the reaction in order to drive its position of equilibrium further to the right. Will this get her a promotion or get her fired? Explain.

The solubility of silver acetate in water at \(20{ }^{\circ} \mathrm{C}\) is \(10.5 \mathrm{~g} / \mathrm{L}\) of solution. Calculate \(\mathrm{K}_{\mathrm{sp}}\) for silver acetate.

In theory, all reactions are reversible, but in practice, some are not. Explain why.

At a given temperature, why is the ratio \(k_{\mathrm{f}} / k_{\mathrm{r}}\) constant for a given reaction?

For the reaction \(4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g)\) at \(25^{\circ} \mathrm{C}, K_{\mathrm{eq}}=0.150 .\) What is the equilibrium concentration of \(\mathrm{NO}_{2}(g)\) if \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.300 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=1.20 \mathrm{M} ?\)
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