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Chapter 14: Problem 67

The partial pressures in an equilibrium mixture of NO, \(\mathrm{Cl}_{2}\), and \(\mathrm{NOCl}\) at \(500 \mathrm{~K}\) are as follows: \(P_{\mathrm{NO}}=0.240 \mathrm{~atm} ;\) \(P_{\mathrm{Cl}_{2}}=0.608 \mathrm{~atm} ; P_{\mathrm{NOCl}}=1.35 \mathrm{~atm}\). What is \(K_{\mathrm{p}}\) at \(500 \mathrm{~K}\) for the reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \operatorname{NOCl}(g) ?\)

Short Answer

Expert verified
The equilibrium constant \(K_{\mathrm{p}}\) is approximately 52.06.

Step by step solution

01

Write the equilibrium expression

For the given reaction \(2 \mathrm{NO}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\), the equilibrium constant \(K_{\mathrm{p}}\) expression is \(K_{\mathrm{p}} = \frac{(P_{\mathrm{NOCl}})^2}{(P_{\mathrm{NO}})^2 \cdot (P_{\mathrm{Cl}_{2}})}\). This represents the ratio of products to reactants at equilibrium, with each partial pressure raised to the power of its coefficient in the balanced chemical equation.
02

Substitute the given partial pressures

Substitute the given partial pressures into the expression for \(K_{\mathrm{p}}\). Using the provided values: \(P_{\mathrm{NO}} = 0.240 \, \mathrm{atm}\), \(P_{\mathrm{Cl}_{2}} = 0.608 \, \mathrm{atm}\), and \(P_{\mathrm{NOCl}} = 1.35 \, \mathrm{atm}\), we have: \(K_{\mathrm{p}} = \frac{(1.35)^2}{(0.240)^2 \cdot (0.608)}\).
03

Calculate \(K_{\mathrm{p}}\)

Perform the calculation: First, square the partial pressures: \((1.35)^2 = 1.8225\), \((0.240)^2 = 0.0576\). Then, use these to find \(K_{\mathrm{p}}: \ K_{\mathrm{p}} = \frac{1.8225}{0.0576 \times 0.608} = \frac{1.8225}{0.0350208} \approx 52.06\).
04

Report \(K_{\mathrm{p}}\)

The calculated equilibrium constant \(K_{\mathrm{p}}\) for the reaction at \(500 \, \mathrm{K}\) is approximately \(52.06\).

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Key Concepts

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

Partial Pressure
Partial pressure refers to the pressure that a single gas in a mixture of gases exerts in its container. Each gas in the mixture contributes to the total pressure. Understanding partial pressure is crucial in chemistry, especially when dealing with gases in reactions.

Consider a mixture of gases like NO, Cl鈧�, and NOCl in a container. Each gas will have its own partial pressure, which adds up to the total pressure in the container.

Partial pressures are important in calculating equilibrium constants, like the one in our exercise. Learning how to calculate and use them can help you understand how gases behave in various conditions.
  • Use the ideal gas law to calculate partial pressures if other properties of the gas are known.
  • The sum of all partial pressures equals the total pressure in the container.
  • Partial pressures are used when calculating the equilibrium constant (Kp).
Equilibrium Expression
The equilibrium expression is a mathematical representation of the balance between the forward and reverse reactions in a chemical system at equilibrium. For gases, the equilibrium constant expression, denoted as \(K_p\), uses partial pressures.

For the reaction given in the exercise:
2 NO(g) + Cl鈧�(g) \(\rightleftharpoons\) 2 NOCl(g), the equilibrium expression can be derived as:
  • Identify the products and reactants in the balanced equation.
  • Write the expression using the formula: \[K_p = \frac{{(P_{\mathrm{NOCl}})^2}}{{(P_{\mathrm{NO}})^2 \cdot (P_{\mathrm{Cl_2}})}}\]
  • The exponents correspond to the coefficients from the balanced equation.
The equilibrium expression shows how Kp depends on the ratio of the pressures, helping to determine the extent of the reaction at equilibrium.
Balanced Chemical Equation
A balanced chemical equation is critical for understanding a chemical reaction completely. It represents the equality in the quantity of moles of reactants and products involved in the reaction.

In chemical equilibrium, it helps to identify the correct stoichiometric coefficients needed to write the equilibrium expression. For example, in our exercise:
2 NO(g) + Cl鈧�(g) \(\rightleftharpoons\) 2 NOCl(g), the equation is already balanced:
  • Each side has 2 nitrogen atoms (from 2 NO) and 1 chlorine molecule (Cl鈧�).
  • Balanced equations maintain the law of conservation of mass.
  • Ensure all elements have the same amount on both sides.
Understanding balanced equations allows us to correctly apply stoichiometry when calculating equilibrium expressions and other related calculations, like determining Kp.
Calculation of Kp
Once you have the equilibrium expression and the partial pressures, you can calculate the equilibrium constant \(K_p\). This constant measures how far a reaction has gone to form products from reactants at equilibrium.

Here鈥檚 how to calculate \(K_p\) for the reaction in our exercise:
Using the values given:
- \(P_{\mathrm{NO}} = 0.240\, \mathrm{atm}\)- \(P_{\mathrm{Cl_2}} = 0.608\, \mathrm{atm}\)- \(P_{\mathrm{NOCl}} = 1.35\, \mathrm{atm}\)
  • Substitute these values into the equilibrium expression: \(K_p = \frac{{(1.35)^2}}{{(0.240)^2 \cdot 0.608}}\).
  • Perform the calculations step-by-step to avoid mistakes.
    • Square the partial pressures as needed.
    • Calculate the denominator.
    • Finally, divide to find \(K_p\).
  • This results in a \(K_p\) of approximately 52.06, illustrating the predominance of NOCl at equilibrium conditions.
Understanding \(K_p\) provides insights into the efficiency and favorability of chemical reactions under given conditions.

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Most popular questions from this chapter

Nitric oxide emitted from the engines of supersonic aircraft can contribute to the destruction of stratospheric ozone: $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \stackrel{k_{\mathrm{L}}}{\rightleftharpoons} \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ This reaction is highly exothermic \((\Delta H=-201 \mathrm{~kJ})\), and its equilibrium constant \(K_{c}\) is \(3.4 \times 10^{34}\) at \(300 \mathrm{~K}\) (a) Which rate constant is larger, \(k_{\mathrm{E}}\) or \(k_{r} ?\) (b) The value of \(k_{\mathrm{f}}\) at \(300 \mathrm{~K}\) is \(8.5 \times 10^{6} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). What is the value of \(k_{r}\) at the same temperature? (c) A typical temperature in the stratosphere is \(230 \mathrm{~K}\). Do the values of \(k_{\mathrm{p}} k_{\mathrm{r}}\) and \(K_{\mathrm{c}}\) increase or decrease when the temperature is lowered from \(300 \mathrm{~K}\) to \(230 \mathrm{~K}\) ?

Consider the equilibrium for the water-gas shift reaction: $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ Use Le Ch氓telier's principle to predict how the concentration of \(\mathrm{H}_{2}\) will change when the equilibrium is disturbed by: (a) Adding \(\mathrm{CO}\) (b) Adding \(\mathrm{CO}_{2}\) (c) Removing \(\mathrm{H}_{2} \mathrm{O}\) (d) Removing \(\mathrm{CO}_{2} ;\) also account for the change using the reaction quotient \(Q_{c}\) -

The following reaction shows the equilibrium reaction that occurs when carbon monoxide enters the blood. (Note: The \(K\) value uses partial pressures of \(\mathrm{O}_{2}\) and \(\mathrm{CO}\) in the equilibrium expression.) \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)(a q)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Hb}(\mathrm{CO})(a q)+\mathrm{O}_{2}(g) \quad K=207\) at \(37^{\circ} \mathrm{C}\) (a) What is the ratio of \([\mathrm{Hb}(\mathrm{CO})]\) to \(\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)\right]\) in air that contains \(20 \% \mathrm{O}_{2}\) and \(0.15 \%\) CO by volume? (b) The treatment for mild carbon monoxide poisoning is breathing pure oxygen. Use Le Ch芒telier's principle to explain why this treatment is effective.

When air is heated at very high temperatures in an automobile engine, the air pollutant nitric oxide is produced by the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad \Delta H^{\circ}=+182.6 \mathrm{~kJ} $$ How does the equilibrium amount of NO vary with an increase in temperature?

Consider the reaction of chloromethane with \(\mathrm{OH}\) in aqueous solution: At \(25^{\circ} \mathrm{C}_{2}\) the rate constant for the forward reaction is \(6 \times 10^{-6} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), and the equilibrium constant \(K_{c}\) is \(1 \times 10^{16}\) Calculate the rate constant for the reverse reaction at \(25^{\circ} \mathrm{C}\).
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