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Chapter 13: Problem 29

For the reaction $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: $[\mathrm{NO}(g)]=8.1 \times 10^{-3} \mathrm{M}\( \)\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{N}_{2}(g)\right]=5.3 \times 10^{-2} M,$ and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]-2.9 \times 10^{-3} \mathrm{M} .\) Calculate the value of \(K\) for the reaction at this temperature.

Short Answer

Expert verified
The value of the equilibrium constant (K) for the given reaction at this temperature is approximately \(K \approx 2.16 \times 10^{5}\).

Step by step solution

01

Write down the balanced chemical equation

Here's the balanced chemical equation for the given reaction: \[2 NO(g)+2 H_{2}(g) \rightleftharpoons N_{2}(g)+2 H_{2}O(g)\]
02

Write down the equilibrium constant expression

According to the law of mass action, the equilibrium constant expression for a given reaction is obtained by multiplying the concentration of the products raised to their stoichiometric coefficients and divided by the multiplication of the concentration of the reactants raised to their stoichiometric coefficients. For the given reaction, the expression for K is: \[K = \frac{[N_{2}]^{1} [H_{2}O]^{2}}{[NO]^{2} [H_{2}]^{2}}\]
03

Substitute the concentrations at equilibrium

We have the equilibrium concentrations as follows: \[[NO(g)] = 8.1 \times 10^{-3} M\] \[[H_{2}(g)] = 4.1 \times 10^{-5} M\] \[[N_{2}(g)] = 5.3 \times 10^{-2} M\] \[[H_{2}O(g)] = 2.9 \times 10^{-3} M\] Substitute these values into the K expression: \[K = \frac{(5.3 \times 10^{-2})^{1} (2.9 \times 10^{-3})^{2}}{(8.1 \times 10^{-3})^{2} (4.1 \times 10^{-5})^{2}}\]
04

Calculate K

Perform the calculations using a calculator to find the value of K at this temperature: \[K \approx \frac{(5.3 \times 10^{-2}) (2.9 \times 10^{-3})^{2}}{(8.1 \times 10^{-3})^{2} (4.1 \times 10^{-5})^{2}} = 2.16 \times 10^{5}\]
05

Write down the final result

The value of the equilibrium constant (K) for the given reaction at this temperature is approximately: \[K \approx 2.16 \times 10^{5}\]

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Most popular questions from this chapter

A 1.00-L flask was filled with 2.00 moles of gaseous \(\mathrm{SO}_{2}\) and 2.00 moles of gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that 1.30 moles of gaseous NO was present. Assume that the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ occurs under these conditions. Calculate the value of the equilibrium constant, \(K,\) for this reaction.

Consider the following reaction at \(725^{\circ} \mathrm{C} :\) $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)$$ At equilibrium, a \(4.50-\mathrm{L}\) container has 2.6 \(\mathrm{g}\) of carbon, \(\mathrm{CO}_{2}\) at a partial pressure of \(0.0020 \mathrm{atm},\) and a total pressure of 0.572 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(725^{\circ} \mathrm{C}\)

Solid \(\mathrm{NH}_{4} \mathrm{HS}\) decomposes by the following endothermic process: $$\mathrm{NH}_{4} \mathrm{HS}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ a. What effect will adding more \(\mathrm{NH}_{3}(g)\) have on the equilibrium? b. What effect will adding more \(\mathrm{NH}_{4} \mathrm{HS}(s)\) have on the equilibrium? c. What effect will increasing the volume of the container have on the equilibrium? d. What effect will decreasing the temperature have on the equilibrium?

Write expressions for \(K_{\mathrm{p}}\) for the following reactions. a. $2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$ b. $\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)$ c. $\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)$ d. $4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)$

The reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ has \(K_{\mathrm{p}}=109\) at \(25^{\circ} \mathrm{C}\) . If the equilibrium partial pressure of \(\mathrm{Br}_{2}\) is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm, calculate the partial pressure of \(\mathrm{NO}\) at equilibrium.
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