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Chapter 12: Problem 66

Consider the equilibrium $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ When this system is at equilibrium at \(700^{\circ} \mathrm{C}\) in a \(2.0\) - \(\mathrm{L}\) container, \(0.10 \mathrm{~mol}\) \(\mathrm{CO}, 0.20 \mathrm{~mol} \mathrm{CO}_{2}\), and \(0.40 \mathrm{~mol} \mathrm{C}\) are present. When the system is cooled to \(600^{\circ} \mathrm{C}\), an additional \(0.040 \mathrm{~mol} \mathrm{C}(s)\) forms. Calculate \(K\) at \(700^{\circ} \mathrm{C}\) and again at \(600^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The equilibrium constant K is 0.025 at 700°C and 0.00125 at 600°C.

Step by step solution

01

Write the balanced chemical equation and the equilibrium expression

We are given the balanced chemical equation: $$\mathrm{C}(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ The equilibrium expression for this reaction is: $$K = \frac{[\mathrm{CO}]^{2}}{[\mathrm{CO}_2 ]}$$
02

Calculate the initial concentrations

We are given the initial moles of the substances in a 2.0 L container. We can calculate the initial concentrations: $$[\mathrm{CO}] = \frac{0.10 \mathrm{~mol}}{2 \mathrm{~L}} = 0.050 \mathrm{~M}$$ $$[\mathrm{CO}_{2}] = \frac{0.20 \mathrm{~mol}}{2 \mathrm{~L}} = 0.10 \mathrm{~M}$$
03

Calculate K at \(700^{\circ} \mathrm{C}\)

We can now substitute the initial concentrations into the equilibrium expression and solve for K: $$K = \frac{[\mathrm{CO}]^{2}}{[\mathrm{CO}_2]} = \frac{(0.050 ^\mathrm{M})^{2}}{0.10 ^\mathrm{M}} = 0.025$$
04

Calculate the new concentrations at \(600^{\circ} \mathrm{C}\)

When the system is cooled to \(600^{\circ} \mathrm{C}\), an additional \(0.040 \mathrm{~mol} \mathrm{C}(s)\) forms. This change will affect the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) as well: Initial moles of \(\mathrm{CO}_{2}\): \(0.20-0.040=0.16\ \mathrm{mol}\) $$[\mathrm{CO}_{2}] = \frac{0.16 \mathrm{~mol}}{2 \mathrm{~L}} = 0.080 \mathrm{~M}$$ Initial moles of \(\mathrm{CO}\): \(0.10-2\times0.040=0.020\ \mathrm{mol}\) $$[\mathrm{CO}] = \frac{0.020 \mathrm{~mol}}{2 \mathrm{~L}} = 0.010 \mathrm{~M}$$
05

Calculate K at \(600^{\circ} \mathrm{C}\)

Using the new concentrations, we can now calculate K for the reaction at \(600^{\circ} \mathrm{C}\) : $$K = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2 ]} = \frac{(0.010 \mathrm{~M})^2}{0.080 \mathrm{~M}} = 0.00125$$ Therefore, the equilibrium constant K is 0.025 at \(700^{\circ} \mathrm{C}\) and 0.00125 at \(600^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

Write equilibrium constant expressions ( \(K\) ) for the following reactions: (a) \(2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \rightleftharpoons\) \(2 \mathrm{NO}(g)+3 \mathrm{Cu}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s)\)

Consider the following reaction at a certain temperature: $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ A reaction mixture contains \(0.70 \mathrm{~atm}\) of \(\mathrm{O}_{2}\) and \(0.81\) atm of NO. When equilibrium is established, the total pressure in the reaction vessel is \(1.20 \mathrm{~atm}\). Find \(K\)

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) \(K=\frac{\left(P_{\mathrm{CO}_{2}}\right)\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)}{\left[\mathrm{CO}_{3}^{2-}\right]\left[\mathrm{H}^{+}\right]^{2}}\)

For the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ \(K\) at a certain temperature is \(0.50\). Predict the direction in which the system will move to reach equilibrium if one starts with (a) \(P_{\mathrm{O}_{2}}=P_{\mathrm{NO}}=P_{\mathrm{NO}_{2}}=0.10 \mathrm{~atm}\) (b) \(P_{\mathrm{NO}_{2}}=0.0848 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.0116 \mathrm{~atm}\) (c) \(P_{\mathrm{NO}_{2}}=0.20 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.010 \mathrm{~atm}, P_{\mathrm{NO}}=0.040 \mathrm{~atm}\)

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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