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Chapter 12: Problem 17

Given the following reactions and their equilibrium constants, $$\begin{array}{lr}\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) & K=1.6 \\ \mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) & K=0.67 \end{array}$$ calculate \(K\) for the reaction$$\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{FeO}(s)+\mathrm{H}_{2}(g)$$

Short Answer

Expert verified
Answer: The equilibrium constant for the given reaction is approximately 2.39.

Step by step solution

01

Manipulate the given reactions

First, we need to reverse the second reaction, so that we have Fe(s) as a reactant: $$\mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{FeO}(s)+\mathrm{CO}(g) \quad K'_{2}=\frac{1}{0.67}$$ Now, we have: $$\begin{array}{l}\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \quad K_{1}=1.6 \\\ \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{FeO}(s)+\mathrm{CO}(g) \quad K'_{2}=\frac{1}{0.67} \end{array}$$
02

Add the manipulated reactions

Now we add the two manipulated reactions together: $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$ $$\mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{FeO}(s)+\mathrm{CO}(g)$$ When summed, we get: $$\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{FeO}(s)+\mathrm{H}_{2}(g)$$
03

Calculate the equilibrium constant for the desired reaction

Since we know the equilibrium constant for the first reaction (\(K_1=1.6\)) and the inverse of the equilibrium constant for the second reaction (\(K'_{2}=\frac{1}{0.67}\)), the equilibrium constant for the desired reaction is the product of those two constants:$$ K_{3} = K_{1} \cdot K'_{2} = 1.6 \cdot \frac{1}{0.67}$$ Now, calculate the product to obtain the value for \(K_3\): $$K_{3} = \frac{1.6}{0.67} \approx 2.39$$ Therefore, the equilibrium constant for the reaction $$\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{FeO}(s)+\mathrm{H}_{2}(g)$$ is \(K_3 \approx 2.39\).

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Most popular questions from this chapter

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?

Derive the relationship $$K=K_{\mathrm{c}} \times(R T)^{\Delta r_{\mathrm{B}}}$$ where \(K_{\mathrm{c}}\) is the equilibrium constant using molarities and \(\Delta n_{\mathrm{g}}\) is the change in the number of moles of gas in the reaction (see page 326). (Hint: Recall that \(P_{\Lambda}=n_{\Lambda} R T / V\) and \(\left.n_{A} / V=[\mathrm{A}] .\right)\)

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 \(25^{\circ} \mathrm{C}\).

Carbonylbromide (COBr_2) can be formed by combining carbon monoxide and bromine gas. $$\mathrm{CO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons \operatorname{COBr}_{2}(g)$$ When equilibrium is established at \(346 \mathrm{~K}\), the partial pressures (in atm) of \(\mathrm{COBr}_{2}\), \(\mathrm{CO}\), and \(\mathrm{Br}_{2}\) are \(0.12,1.00\), and \(0.65\), respectively. (a) What is \(K\) at \(346 \mathrm{~K} ?\) (b) Enough bromine condenses to decrease its partial pressure to \(0.50\) atm. What are the equilibrium partial pressures of all gases after equilibrium is re-established?

Hydrogen cyanide, a highly toxic gas, can decompose to cyanogen and hydrogen gases, $$2 \mathrm{HCN}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{~N}_{2}(g)+\mathrm{H}_{2}(g)$$ At a certain temperature, \(K\) for this decomposition is \(0.17\). What are the partial pressures of all gases at equilibrium if initially the partial pressures are \(P_{\mathrm{C}_{2} \mathrm{~N}_{2}}=P_{\mathrm{H}_{2}}=0.32 \mathrm{~atm}, P_{\mathrm{HCN}}=0.45 \mathrm{~atm} ?\)
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