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Chapter 12: Problem 31

A compound, \(\mathrm{X}\), decomposes at \(131^{\circ} \mathrm{C}\) according to the following equation: $$2 \mathrm{X}(g) \rightleftharpoons \mathrm{A}(g)+3 \mathrm{C}(g) \quad K=1.1 \times 10^{-3}$$ If a flask initially contains \(\mathrm{X}, \mathrm{A}\), and \(\mathrm{C}\), all at partial pressures of \(0.250 \mathrm{~atm}\), in which direction will the reaction proceed?

Short Answer

Expert verified
Answer: The reaction will proceed in the reverse direction to reach equilibrium.

Step by step solution

01

Understand the reaction and write the equilibrium constant expression.

Given the reaction: $$2 \mathrm{X}(g) \rightleftharpoons \mathrm{A}(g)+3 \mathrm{C}(g) \quad K=1.1 \times 10^{-3}$$ We can write the equilibrium constant expression as: $$K = \frac{[\mathrm{A}][\mathrm{C}]^3}{[\mathrm{X}]^2}$$
02

Calculate the initial reaction quotient (Q).

Using the initial partial pressures of X, A, and C, we can calculate the initial reaction quotient: $$Q = \frac{[\mathrm{A}][\mathrm{C}]^3}{[\mathrm{X}]^2} = \frac{(0.250)(0.250)^3}{(0.250)^2}$$
03

Simplify the expression for the reaction quotient (Q).

After simplification, we have: $$Q=\frac{(0.250)^4}{(0.250)^2} = (0.250)^2$$
04

Evaluate the reaction quotient (Q) and compare to the equilibrium constant (K).

We have: $$Q = (0.250)^2 = 0.0625$$ Now compare Q to K: $$Q > K \Rightarrow (0.0625 > 1.1 \times 10^{-3})$$
05

Determine the direction of the reaction.

Since Q is greater than K, the reaction will proceed in the reverse direction to reach the equilibrium. The decomposition of X will decrease, and the formation of A and C will increase until Q equals to K, reaching the equilibrium.

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Most popular questions from this chapter

Write a chemical equation for an equilibrium system that would lead to the following expressions (a-d) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{CO}_{2}}\right)^{3}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{4}}{\left(P_{\mathrm{C}, \mathrm{H}_{4}}\right)\left(P_{\mathrm{O}_{2}}\right)^{5}}\) (b) \(K=\frac{P_{\mathrm{C}_{0} \mathrm{H}_{12}}}{\left(P_{\mathrm{C}, \mathrm{H}_{6}}\right)^{2}}\) (c) \(K=\frac{\left[\mathrm{PO}_{4}^{3-}\right]\left[\mathrm{H}^{+}\right]^{3}}{\left[\mathrm{H}_{3} \mathrm{PO}_{4}\right]}\) (d) \(K=\frac{\left(P_{\mathrm{CO}_{2}}\right)\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)}{\left[\mathrm{CO}_{3}^{2-}\right]\left[\mathrm{H}^{+}\right]^{2}}\)

Ammonium carbamate solid \(\left(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\right)\) decomposes at \(313 \mathrm{~K}\) into ammonia and carbon dioxide gases. At equilibrium, analysis shows that there are \(0.0451\) atm of \(\mathrm{CO}_{2}, 0.0961\) atm of ammonia, and \(0.159 \mathrm{~g}\) of ammonium carbamate. (a) Write a balanced equation for the decomposition of one mole of \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\) (b) Calculate \(K\) at \(313 \mathrm{~K}\).

A sealed flask has \(0.541\) atm of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\). The following equilibrium is established. $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium, the partial pressure of oxygen is measured to be \(0.216\) atm. Calculate \(K\) for the decomposition of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\).

. At \(800 \mathrm{~K}\), hydrogen iodide can decompose into hydrogen and iodine gases. $$2 \mathrm{HI}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{H}_{2}(g)$$ At this temperature, \(K=0.0169 .\) What are the partial pressures at equilibrium of the hydrogen and iodine if initially a sealed flask at \(800 \mathrm{~K}\) contains only HI at a pressure of \(0.200\) atm?

Given the following data at \(25^{\circ} \mathrm{C}\), $$\begin{array}{cl}2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K=1 \times 10^{-30} \\ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) & K=8 \times 10^{1} \end{array}$$ calculate \(K\) for the formation of one mole of NOBr from its elements in the gaseous state at \(25^{\circ} \mathrm{C}\).
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