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Chapter 15: Problem 2

Based on these descriptions, write a balanced equation and the corresponding \(K_{\mathrm{p}}\) expression for each reversible reaction. (a) Oxygen gas oxidizes gaseous ammonia to gaseous nitrogen and water vapor. (b) Hydrogen gas reduces gaseous nitrogen dioxide to gaseous ammonia and water vapor. (c) Nitrogen gas reacts with the solid sodium carbonate and carbon to produce solid sodium cyanide and carbon monoxide gas.

Short Answer

Expert verified
The balanced chemical equations are: \n(a) \(4NH_3 + 5O_2 \rightarrow 4N_2 + 6H_2O\) with \(K_{\mathrm{p}} = \frac{(P_{N2})^4 (P_{H2O})^6}{(P_{NH3})^4 (P_{O2})^5}\) \n(b) \(3H_2 + 2NO_2 \rightarrow 2NH_3 + 2H_2O\) with \(K_{\mathrm{p}} = \frac{(P_{NH3})^2 (P_{H2O})^2}{(P_{H2})^3 (P_{NO2})^2}\) \n(c) \(N_2 + 4Na_2CO_3 + C \rightarrow 4NaCN + 3CO\) with \(K_{\mathrm{p}} = (P_{CO})^3/\(P_{N2} \(P_{C}\)

Step by step solution

01

Problem a - Balancing Chemical Equation

First, let's write down the unbalanced equation using the given elements and compounds: \(O_2 + NH_3 \rightarrow N_2 + H_2O\). Now let's balance this equation. The balanced chemical equation is: \(4NH_3 + 5O_2 \rightarrow 4N_2 + 6H_2O\).
02

Problem a - \(K_{\mathrm{p}}\) expression

The \(K_{\mathrm{p}}\) expression for a reaction at equilibrium is the ratio of the products to the reactants, each raised to the power of its stoichiometric coefficient in the balanced chemical equation. Therefore, the \(K_{\mathrm{p}}\) expression for this reaction is: \(K_{\mathrm{p}} = \frac{(P_{N2})^4 (P_{H2O})^6}{(P_{NH3})^4 (P_{O2})^5}\).
03

Problem b - Balancing Chemical Equation

Similarly, let’s write down the unbalanced equation for the reaction: \(H_2 + NO_2 \rightarrow NH_3 + H_2O\). Then, we balance the equation, which gives: \(3H_2 + 2NO_2 \rightarrow 2NH_3 + 2H_2O\).
04

Problem b - \(K_{\mathrm{p}}\) expression

Following the same process as in part a, the \(K_{\mathrm{p}}\) expression for this reaction is: \(K_{\mathrm{p}} = \frac{(P_{NH3})^2 (P_{H2O})^2}{(P_{H2})^3 (P_{NO2})^2}\).
05

Problem c - Balancing Chemical Equation

Let’s write down the unbalanced equation for the reaction: \(N_2 + Na_2CO_3 + C \rightarrow NaCN + CO\). Then, balance this equation, which results in: \(N_2 + 4Na_2CO_3 + C \rightarrow 4NaCN + 3CO\).
06

Problem c - \(K_{\mathrm{p}}\) expression

The \(K_{\mathrm{p}}\) expression is a bit different in this case because a solid (Na_2CO_3 and NaCN) is involved. The concentrations of solids are not included in the equilibrium expressions, so the \(K_{\mathrm{p}}\) expression for this reaction is: \(K_{\mathrm{p}} = (P_{CO})^3/\(P_{N2} \(P_{C}\).

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Most popular questions from this chapter

Determine \(K_{c}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})+\) \(\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g}),\) given the following data at \(298 \mathrm{K}\) $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.0 \times 10^{-9}$$ $$\operatorname{NOCl}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=1.1 \times 10^{2}$$ $$\mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=0.3$$

In the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}), 0.455\) \(\mathrm{mol} \mathrm{SO}_{2}, 0.183 \mathrm{mol} \mathrm{O}_{2},\) and \(0.568 \mathrm{mol} \mathrm{SO}_{3}\) are introduced simultaneously into a 1.90 L vessel at \(1000 \mathrm{K}\). (a) If \(K_{c}=2.8 \times 10^{2},\) is this mixture at equilibrium? (b) If not, in which direction will a net change occur?

Equilibrium is established in the reversible reaction \(2 \mathrm{A}+\mathrm{B} \rightleftharpoons 2 \mathrm{C} .\) The equilibrium concentrations are \([\mathrm{A}]=0.55 \mathrm{M},[\mathrm{B}]=0.33 \mathrm{M},[\mathrm{C}]=0.43 \mathrm{M}\) What is the value of \(K_{c}\) for this reaction?

In the human body, the enzyme carbonic anahydrase catalyzes the interconversion of \(\mathrm{CO}_{2}\) and \(\mathrm{HCO}_{3}^{-}\) by either adding or removing the hydroxide anion. The overall reaction is endothermic. Explain how the following affect the amount of carbon dioxide: (a) increasing the amount of bicarbonate anion; (b) increasing the pressure of carbon dioxide; (c) increasing the amount of carbonic anhydrase; (d) decreasing the temperature.

Refer to Example \(15-4 . \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) at \(747.6 \mathrm{mmHg}\) pressure and a \(1.85 \mathrm{g}\) sample of \(\mathrm{I}_{2}(\mathrm{s})\) are introduced into a \(725 \mathrm{mL}\) flask at \(60^{\circ} \mathrm{C} .\) What will be the total pressure in the flask at equilibrium? $$\begin{aligned} \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{s}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})+\mathrm{S}(\mathrm{s}) & \\ K_{\mathrm{p}}=& 1.34 \times 10^{-5} \mathrm{at} 60^{\circ} \mathrm{C} \end{aligned}$$
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