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Chapter 12: Problem 22

Calculate \(K\) for the formation of methyl alcohol at \(100^{\circ} \mathrm{C}\) : $$\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ given that at equilibrium, the partial pressures of the gases are \(P_{\mathrm{CO}}=0.814 \mathrm{~atm}, P_{\mathrm{H}_{2}}=0.274 \mathrm{~atm}\), and \(P_{\mathrm{CH}_{3} \mathrm{OH}}=0.0512 \mathrm{~atm} .\)

Short Answer

Expert verified
Question: Calculate the equilibrium constant, K, for the formation of methyl alcohol (CH3OH) at 100°C given the partial pressures of CO, H2, and CH3OH at equilibrium as 0.814 atm, 0.274 atm, and 0.0512 atm, respectively. Answer: The equilibrium constant K for the formation of methyl alcohol at 100°C is 1.24.

Step by step solution

01

Write the expression for K_p

For the given balanced chemical equation, we can write the expression for the equilibrium constant K_p as: $$K_p = \frac{P_{\mathrm{CH}_{3} \mathrm{OH}}}{P_{\mathrm{CO}} \cdot P_{\mathrm{H}_{2}}^2}$$
02

Substitute the given values

We can now substitute the given values of partial pressures at equilibrium into the expression for K_p: $$K_p = \frac{0.0512\,\mathrm{atm}}{(0.814\,\mathrm{atm}) \cdot (0.274\,\mathrm{atm})^2}$$
03

Calculate K_p

Now, we can calculate K_p by performing the calculation: $$K_p = \frac{0.0512}{(0.814) \cdot (0.274)^2} = 1.24$$ So, the equilibrium constant K for the formation of methyl alcohol at 100°C is 1.24.

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Most popular questions from this chapter

Carbonyl fluoride, \(\mathrm{COF}_{2}\), is an important intermediate for organic fluorine compounds. It can be prepared by the following reaction: $$\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{CF}_{4}(g) \rightleftharpoons 2 \mathrm{COF}_{2}(g)$$ At \(1000^{\circ} \mathrm{C}, K\) for this reaction is \(0.50 .\) What are the partial pressures of all the gases at equilibrium when the initial partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CF}_{4}\) are \(0.713 \mathrm{~atm} ?\)

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is \(0.45\). If \(20.0 \mathrm{~mL}\) \((d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00\) -L vessel and heated to \(180^{\circ} \mathrm{C}\), what percentage remains undissociated at equilibrium?

WEB Write equilibrium constant \((K)\) expressions for the following reactions: (a) \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\)

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 \(K\)

. 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?
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