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Chapter 9: Problem 84

A mixture consisting of \(1.000 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) and \(1.000 \mathrm{~mol} \mathrm{CO}(\mathrm{g})\) is placed in a \(10.00-\mathrm{L}\) reaction vessel at \(800 \mathrm{~K}\). At equilibrium, \(0.665 \mathrm{~mol} \mathrm{CO}_{2}(\mathrm{~g})\) is present as a result of the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\). What are (a) the equilibrium concentrations for all substances and (b) the value of \(K_{e}\) ?

Short Answer

Expert verified
The equilibrium concentrations are: \( [CO] = [H_2O] = 0.0335 \text{ M} \), \( [CO_2] = [H_2] = 0.0665 \text{ M} \). The value of \( K_c \) is approximately 3.94.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the reaction is already given: \( \text{CO}(g) + \text{H}_2\text{O}(g) \rightarrow \text{CO}_2(g) + \text{H}_2(g) \). This reaction is at equilibrium.
02

Determine the initial concentrations

Calculate the initial concentrations (in mol/L) of the reactants using the provided initial amounts and the volume of the reaction vessel. For \( \text{CO}(g) \) and \( \text{H}_2\text{O}(g) \), both initially have 1.000 mol in a 10.00 L container, so their initial concentrations are both \( 1.000 \text{ mol} / 10.00 \text{ L} = 0.1000 \text{ M} \). Since \( \text{CO}_2(g) \) and \( \text{H}_2(g) \) are products of the reaction and not initially present, their initial concentrations are zero.
03

Determine the change in concentrations

At equilibrium, 0.665 mol of \( \text{CO}_2 \) is present, so the same amount of \( \text{H}_2 \) will also be present due to the 1:1 stoichiometry. Since both \( \text{CO} \) and \( \text{H}_2\text{O} \) decrease by 0.665 mol to produce \( \text{CO}_2 \) and \( \text{H}_2 \), calculate the decrease in concentration as \( 0.665 \text{ mol} / 10.00 \text{ L} = 0.0665 \text{ M} \).
04

Calculate equilibrium concentrations

Subtract the change in concentration from the initial concentrations to find the equilibrium concentrations for the reactants and add the change to determine the concentrations for the products. For \( \text{CO} \) and \( \text{H}_2\text{O} \): \( 0.1000 \text{ M} - 0.0665 \text{ M} = 0.0335 \text{ M} \). For \( \text{CO}_2 \) and \( \text{H}_2 \): \( 0 \text{ M} + 0.0665 \text{ M} = 0.0665 \text{ M} \).
05

Calculate the equilibrium constant \( K_c \)

The equilibrium constant \( K_c \) is calculated using the equilibrium concentrations of the products and reactants raised to the power of their stoichiometric coefficients: \( K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{(0.0665)(0.0665)}{(0.0335)(0.0335)} \). Compute this value to find the equilibrium constant.
06

Determine the equilibrium constant \( K_c \)

Perform the calculation from the previous step to find the value of \( K_c \). \( K_c = \frac{(0.0665)(0.0665)}{(0.0335)(0.0335)} = \frac{0.00442225}{0.00112225} \approx 3.94 \).

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Key Concepts

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

Chemical Reaction
A chemical reaction involves the transformation of reactants into products. This process occurs when reactant molecules collide with sufficient energy to overcome the activation energy barrier, leading to a rearrangement of atoms to form new substances. In our example, the chemical reaction under study is a combination of carbon monoxide (CO) and water vapor (H鈧侽) to produce carbon dioxide (CO鈧�) and hydrogen gas (H鈧�).

This transformation honors the law of conservation of mass, meaning that the number of atoms of each element remains the same before and after the reaction. To visualize this concept, picture a dance floor where dance partners (atoms) switch partners (bonds) without leaving the dance (reactor). This particular dance is orchestrated so that each dancer (atom) ends up with a new partner (molecule) in a 1:1 ratio, following the stoichiometry of the reaction.
Equilibrium Constant
In the realm of chemistry, when a reaction reaches a stage where the rate of the forward reaction equals the rate of the reverse reaction, it is said to be at equilibrium. The equilibrium constant, symbolized by Kc, is a numerical value that quantifies the ratio of concentrations of products to reactants at this state, each raised to the power of their stoichiometric coefficients.

Picture the balance scale; the equilibrium constant is like the scale's reading, illustrating whether the reaction favors products, reactants, or is balanced. A larger Kc suggests products are favored, indicating a 'heavier' side with more products, while a smaller Kc implies the scale tips towards the reactants, suggesting their dominance at equilibrium.
Stoichiometry
The art of stoichiometry lies in the quantitative relationship between reactants and products in a chemical reaction. It's like a recipe that dictates the exact proportions of ingredients needed and the amount of product formed.

In our scenario, the stoichiometry of the reaction is 1:1:1:1, signifying for every mole of CO and H鈧侽 that reacts, one mole of CO鈧� and H鈧� will form. Furthermore, when 0.665 mol of CO鈧� is produced, an equal amount of H鈧� forms while the same quantity of CO and H鈧侽 is consumed. This stoichiometric relationship is crucial for calculating the changes in concentrations and is a fundamental concept in understanding chemical reactions.
Equilibrium Concentrations
The equilibrium concentrations refer to the amounts of reactants and products present when the chemical reaction has reached equilibrium. These concentrations can be calculated systematically, often starting with initial concentrations and then adjusting based on the reaction stoichiometry and changes in moles of substances as the reaction progresses.

In the given exercise, we applied stoichiometry to find the changes in concentrations from initial to equilibrium. Like a bookkeeper, we 'debited' the consumed reactants and 'credited' the formed products to their respective 'accounts' to tally the equilibrium 'balance'. The concentration of each substance at equilibrium represents its 'final balance', which plays a pivotal role in calculating the equilibrium constant.

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Most popular questions from this chapter

Write the equilibrium expressions \(K_{c}\) for the following reactions. (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}(\mathrm{g})+\mathrm{Cl}(\mathrm{g})\) (b) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g})\) (c) \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Write the reaction quotients \(Q_{c}\) for (a) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CuCl}_{2}(\mathrm{~s})\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) (s) \(\rightarrow \mathrm{N}_{2} \mathrm{O}\) (g) \(+2 \mathrm{H}_{2} \mathrm{O}\) (g) (c) \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{MgO}\) (s) \(+\mathrm{CO}_{2}\) (g)

\( \mathrm{~A} 25.0 \mathrm{~g}\) sample of ammonium carbamate, \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)\), was placed in an evacuated \(0.250-\mathrm{L}\). flask and kept at \(25^{\circ} \mathrm{C}\). At equilibrium, the flask contained \(17.4 \mathrm{mg}\) of \(\mathrm{CO}_{2}\). What is the value of \(K_{c}\) for the decomposition of ammonium carbamate into ammonia and carbon dioxide? The reaction is \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(\mathrm{s})=2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) .\)

(a) Calculate the reaction free energy of \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Hl}(\mathrm{g})\) at \(700 \mathrm{~K}\) when the concentrations of \(\mathrm{H}_{2}, \mathrm{I}_{2}\), and HI are \(0.026,0.33\), and \(1.84 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), respectively. For this reaction, \(K_{c}=54\) at \(700 \mathrm{~K}\). (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

Let \(\alpha\) be the fraction of \(\mathrm{PCl}_{5}\) molecules that have decomposed to \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) in the reaction \(\mathrm{PCl}_{5}(\mathrm{~g})=\) \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) in a constant- volume container, so that the amount of \(\mathrm{PCl}_{5}\) at equilibrium is \(n(1-\alpha)\), where \(n\) is the pressure present initially. Derive an equation for \(K\) in terms of \(\alpha\) and the total pressure \(P\), and solve it for \(\alpha\) in terms of \(P\). Calculate the fraction decomposed at \(556 \mathrm{~K}\), at which temperature \(K=4.96\), and the total pressure is (a) \(0.50\) har; (b) \(1.00\) bar.
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