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Chapter 12: Problem 21

When carbon monoxide reacts with hydrogen gas, methane and steam are formed. $$\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ At \(1127^{\circ} \mathrm{C}\), analysis at equilibrium shows that \(P_{\mathrm{CO}}=0.921 \mathrm{~atm}\), \(P_{\mathrm{H}_{2}}=1.21 \mathrm{~atm}, P_{\mathrm{CH}_{4}}=0.0391 \mathrm{~atm}\), and \(P_{\mathrm{H}_{2} \mathrm{O}}=0.0124 \mathrm{~atm} .\) What is the equilibrium constant, \(K\), for the reaction at \(1127^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
Question: Calculate the equilibrium constant (K) for the given reaction at 1127°C, given the partial pressures at equilibrium as P_CO = 0.921 atm, P_H2 = 1.21 atm, P_CH4 = 0.0391 atm, and P_H2O = 0.0124 atm: CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g)

Step by step solution

01

Write down the equilibrium constant expression for the given reaction

The equilibrium constant (K) for the reaction is given by the expression, K = (P_{CH4} * P_{H2O}) / (P_{CO} * (P_{H2})^3) This expression is derived from the balanced chemical equation given, using the Law of Mass Action.
02

Substitute the given values

We are given that at equilibrium: P_{CO}=0.921 atm, P_{H2}=1.21 atm, P_{CH4}=0.0391 atm, and P_{H2O}=0.0124 atm. Substitute these values into the equilibrium constant expression. K = (0.0391 * 0.0124) / (0.921 * (1.21)^3)
03

Calculate the equilibrium constant, K

Now it's a matter of performing the calculation. Don't forget to raise the partial pressure of hydrogen gas to the power of 3 as per its stoichiometric coefficient in the balanced chemical equation K = 0.00048504 / (0.921 * 1.770651) And when you calculate it further, K = 0.000247 This is the equilibrium constant for the reaction at 1127°C.

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Most popular questions from this chapter

At a certain temperature, \(K\) is \(1.3 \times 10^{5}\) for the reaction $$2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g)$$ What is the equilibrium pressure of hydrogen sulfide if those of hydrogen and sulfur gases are \(0.103\) atm and \(0.417\) atm, respectively?

Consider the statement "The equilibrium constant for a reaction at \(400 \mathrm{~K}\) is 792 . It must be a very fast reaction." What is wrong with the statement?

At \(25^{\circ} \mathrm{C}, K=2.2 \times 10^{-3}\) for the reaction $$\mathrm{ICl}(g) \rightleftharpoons \frac{1}{2} \mathrm{I}_{2}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g)$$ Calculate \(K\) at \(25^{\circ} \mathrm{C}\) for (a) the decomposition of ICl into one mole of iodine and chlorine. (b) the formation of two moles of \(\operatorname{ICl}(g)\).

Benzaldehyde, a flavoring agent, is obtained by the dehydrogenation of benzyl alcohol. $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(g)+\mathrm{H}_{2}(g)$$ \(K\) for the reaction at \(250^{\circ} \mathrm{C}\) is \(0.56\). If \(1.50 \mathrm{~g}\) of benzyl alcohol is placed in a 2.0-L flask and heated to \(250^{\circ} \mathrm{C}\), (a) what is the partial pressure of the benzaldehyde when equilibrium is established? (b) how many grams of benzyl alcohol remain at equilibrium?

At a certain temperature, \(K\) is \(0.040\) for the decomposition of two moles of bromine chloride gas (BrCl) to its elements. An equilibrium mixture at this temperature contains bromine and chlorine gases at equal partial pressures of \(0.0493\) atm. What is the equilibrium partial pressure of bromine chloride?
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