The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal: $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ At a given temperature, \(K_{\mathrm{p}}=2.7 .\) If \(0.13 \mathrm{~mol}\) of \(\mathrm{CO}, 0.56 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}, 0.62 \mathrm{~mol}\) of \(\mathrm{CO}_{2},\) and \(0.43 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) are put in a \(2.0-\mathrm{L}\) flask, in which direction does the reaction proceed?

Short Answer

Expert verified
The reaction will proceed in the reverse direction.

Step by step solution

01

Write the Reaction and Given Data

The balanced chemical equation for the water-gas shift reaction is \(\mathrm{CO}(g) + \mathrm{H}_{2}\mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g) + \mathrm{H}_{2}(g)\). The equilibrium constant, \(K_{\mathrm{p}}\), at a given temperature is 2.7. The initial moles of each component are as follows:\[\text{CO} = 0.13 \; \text{mol}\]\[\text{H}_2\text{O} = 0.56 \; \text{mol}\]\[\text{CO}_2 = 0.62 \; \text{mol}\]\[\text{H}_2 = 0.43 \; \text{mol}\]\
02

Calculate Initial Concentrations

Convert the moles of each gas to concentrations using the volume of the flask, \(2.0 \; \text{L}\).\[\left[ \text{CO} \right] = \frac{0.13 \; \text{mol}}{2.0 \; \text{L}} = 0.065 \; \text{M}\]\[\left[ \text{H}_2\text{O} \right] = \frac{0.56 \; \text{mol}}{2.0 \; \text{L}} = 0.28 \; \text{M}\]\[\left[ \text{CO}_2 \right] = \frac{0.62 \; \text{mol}}{2.0 \; \text{L}} = 0.31 \; \text{M}\]\[\left[ \text{H}_2 \right] = \frac{0.43 \; \text{mol}}{2.0 \; \text{L}} = 0.215 \; \text{M}\]\
03

Write the Reaction Quotient (Qp) Expression

The reaction quotient, \(Q_{\mathrm{p}}\), is given by:\[ Q_{\mathrm{p}} = \frac{ \left[ \mathrm{CO}_{2} \right] \left[ \mathrm{H}_{2} \right] }{ \left[ \mathrm{CO} \right] \left[ \mathrm{H}_2\mathrm{O} \right] }\]\ Substitute the initial concentrations into the expression:\[ Q_{\mathrm{p}} = \frac{ (0.31)(0.215) }{ (0.065)(0.28) } = \frac{0.06665}{0.0182} \approx 3.66\]
04

Compare Qp to Kp

Compare the calculated \(Q_{\mathrm{p}}\) to the given \(K_{\mathrm{p}} = 2.7\). Since \(Q_{\mathrm{p}} = 3.66\) is greater than \(K_{\mathrm{p}} = 2.7\), the reaction will shift to the left to reach equilibrium.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

equilibrium constant
The equilibrium constant, denoted as \( K \), is a crucial concept in chemical reactions. It gives us an idea of the ratio of concentrations of products to reactants at equilibrium. When we have a reaction:
\(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant \(K\) for this reaction can be written as:
\[ K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]The square brackets indicate the concentration of each species, and the exponents are the stoichiometric coefficients from the balanced equation.
Higher values of \( K \) mean more products than reactants at equilibrium. Conversely, lower values of \( K \) mean more reactants than products.Understanding \( K \) helps us predict the position of equilibrium and the extent of the reaction.
In the water-gas shift reaction, we are given \( K_p = 2.7 \). This value helps us in determining if the system is at equilibrium or if it will shift to reactants or products to reach equilibrium.
reaction quotient
The reaction quotient, \(Q\), provides a snapshot of a reaction's progress. While \(K\) uses equilibrium concentrations, \(Q\) uses the current, known concentrations at any point in time.
For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the reaction quotient \(Q\) can be calculated similarly:\[ Q = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]
By comparing \(Q\) with the equilibrium constant \(K\), we can predict the direction the reaction needs to proceed to reach equilibrium:
  • If \(Q < K\), the reaction will shift to the right, producing more products.
  • If \(Q > K\), the reaction will shift to the left, forming more reactants.
  • If \(Q = K\), the reaction is at equilibrium.
In our example with the water-gas shift reaction: \[ Q_p = \frac{(0.31)(0.215)}{(0.065)(0.28)} \approx 3.66 \], which we then compare with \( K_p = 2.7 \).
direction of reaction
The direction in which a reaction proceeds depends on the relationship between the reaction quotient, \(Q\), and the equilibrium constant, \(K\).
If \(Q < K\), the system has too many reactants compared to products, so the reaction will move forward (to the right) to produce more products. Conversely, if \(Q > K\), there are too many products, and the reaction will go in reverse (to the left) to produce more reactants.
In our specific example, \( Q_p = 3.66 \), and \( K_p = 2.7 \). Since \( Q_p \) is greater than \( K_p \), the reaction favors the reactants. Thus, the system will shift to the left.
Understanding the direction of a reaction based on \( Q \) and \( K \) helps chemists control and optimize reactions, essential in various industrial and laboratory processes.

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Most popular questions from this chapter

In a study of the formation of HI from its elements, $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ equal amounts of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) were placed in a container, which was then sealed and heated. (a) On one set of axes, sketch concentration vs. time curves for \(\mathrm{H}_{2}\) and HI, and explain how \(Q\) changes as a function of time. (b) Is the value of \(Q\) different if \(\left[\mathrm{I}_{2}\right]\) is plotted instead of \(\left[\mathrm{H}_{2}\right] ?\)

Isolation of Group \(8 \mathrm{~B}(10)\) elements, used as industrial catalysts, involves a series of steps. For nickel, the sulfide ore is roasted in air: \(\mathrm{Ni}_{3} \mathrm{~S}_{2}(s)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{NiO}(s)+\mathrm{SO}_{2}(g) .\) The metal oxide is reduced by the \(\mathrm{H}_{2}\) in water gas \(\left(\mathrm{CO}+\mathrm{H}_{2}\right)\) to impure \(\mathrm{Ni}: \mathrm{NiO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{H}_{2} \mathrm{O}(g) .\) The \(\mathrm{CO}\) in water gas then reacts with the metal in the Mond process to form gaseous nickel carbonyl, \(\mathrm{Ni}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g),\) which is sub- sequently decomposed to the metal. (a) Balance each of the three steps, and obtain an overall balanced equation for the conversion of \(\mathrm{Ni}_{3} \mathrm{~S}_{2}\) to \(\mathrm{Ni}(\mathrm{CO})_{4}\). (b) Show that the overall \(Q_{\mathrm{c}}\) is the product of the \(Q_{c}\) 's for the individual reactions.

Determine \(\Delta n_{\text {gas }}\) for each of the following reactions: (a) \(\mathrm{MgCO}_{3}(s) \rightleftharpoons \mathrm{MgO}(s)+\mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{HNO}_{3}(l)+\mathrm{ClF}(g) \rightleftharpoons \mathrm{ClONO}_{2}(g)+\mathrm{HF}(g)\)

Predict the effect of increasing the container volume on the amounts of each reactant and product in the following reactions: (a) \(\mathrm{CH}_{3} \mathrm{OH}(l) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\) (b) \(\mathrm{CH}_{4}(g)+\mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{HCN}(g)+3 \mathrm{H}_{2}(g)\)

Hydrogen sulfide decomposes according to the following reaction, for which \(K_{c}=9.30 \times 10^{-8}\) at \(700^{\circ} \mathrm{C}:\) $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ If \(0.45 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{~S}\) is placed in a 3.0 - \(\mathrm{L}\) container, what is the equilibrium concentration of \(\mathrm{H}_{2}(g)\) at \(700^{\circ} \mathrm{C} ?\)

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