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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

. Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ At a certain temperature, the equilibrium constant for the reaction is \(0.0255\). What are the partial pressures of all gases at equilibrium if the initial partial pressure of the gases (both reactants and products) is \(0.300 \mathrm{~atm} ?\)

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 \(1000 \mathrm{~K}\).

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 \(\left.n_{A} / V=[\mathrm{A}] .\right)\)

Solid ammonium iodide decomposes to ammonia and hydrogen iodide gases at sufficiently high temperatures. $$\mathrm{NH}_{4} \mathrm{I}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HI}(g)$$ The equilibrium constant for the decomposition at \(673 \mathrm{~K}\) is \(0.215\). Fifteen grams of ammonium iodide are sealed in a \(5.0\) -L flask and heated to \(673 \mathrm{~K}\). (a) What is the total pressure in the flask at equilibrium? (b) How much ammonium iodide decomposes?

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 \(313 \mathrm{~K}\).
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