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Chapter 19: Problem 32

Given that (1) \(\mathrm{C}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \leftrightharpoons \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2}(g)\) \(K_{\mathrm{P}_{1}}=3.85\) bar (2) \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \leftrightharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g)\) \(K_{\mathrm{P}_{2}}=0.71\) calculate the value of \(K_{\mathrm{p}}\) for the equation (3) \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)\)

Short Answer

Expert verified
The value of \( K_{P} \) for reaction (3) is 0.184.

Step by step solution

01

Analyze Given Reactions

We have two reactions and their equilibrium constants, denoted by \( K_{P_1} \) and \( K_{P_2} \). Reaction (1) is \( \mathrm{C}(s)+2 \mathrm{H}_2 \mathrm{O}(g) \leftrightharpoons \mathrm{CO}_2(g)+2 \mathrm{H}_2(g) \) with \( K_{P_1} = 3.85 \), and reaction (2) is \( \mathrm{H}_2(g)+\mathrm{CO}_2(g) \leftrightharpoons \mathrm{H}_2 \mathrm{O}(g)+\mathrm{CO}(g) \) with \( K_{P_2} = 0.71 \). Our goal is to determine \( K_{P} \) for reaction (3): \( \mathrm{C}(s)+\mathrm{CO}_2(g) \leftrightharpoons 2 \mathrm{CO}(g) \).
02

Understand Reaction Combination

To derive the equilibrium expression for the desired reaction (3), we need to combine reactions (1) and (2). By manipulating these reactions, we will equate it to the desired reaction.
03

Manipulate Reaction (1)

To help derive reaction (3), reverse reaction (1). This gives us \( \mathrm{CO}_2(g)+2 \mathrm{H}_2(g) \leftrightharpoons \mathrm{C}(s)+2 \mathrm{H}_2 \mathrm{O}(g) \). The equilibrium constant for the reverse reaction is the reciprocal: \( \frac{1}{K_{P_1}} = \frac{1}{3.85} \).
04

Combine Reactions to Match Desired Reaction

Add Reaction (2) as given \( \mathrm{H}_2(g)+\mathrm{CO}_2(g) \leftrightharpoons \mathrm{H}_2 \mathrm{O}(g)+\mathrm{CO}(g) \) to the reversed Reaction (1). This will eliminate \( \mathrm{H}_2 \) and \( \mathrm{H}_2 \mathrm{O} \), yielding: \[\mathrm{CO}_2(g) + 2\mathrm{H}_2(g) + \mathrm{H}_2(g) +\mathrm{CO}_2(g) \leftrightharpoons \mathrm{C}(s)+2 \mathrm{H}_2 \mathrm{O}(g) + \mathrm{H}_2\mathrm{O}(g) + \mathrm{CO}(g)\]Combine terms to simplify to: \[\mathrm{C}(s) + \mathrm{CO}_2(g) \leftrightharpoons 2 \mathrm{CO}(g)\].This matches reaction (3).
05

Calculate Combined Equilibrium Constant

The equilibrium constant \( K_{P} \) for the overall reaction (desired reaction 3) is the product of the manipulated constants: \[ K_{P} = \frac{1}{K_{P_1}} \times K_{P_2} \].Substitute the values: \[ K_{P} = \left(\frac{1}{3.85}\right) \times 0.71 \].
06

Perform Final Calculation

Calculate the numerical value: \[ K_{P} = \left(\frac{1}{3.85}\right) \times 0.71 = 0.184 \].

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Key Concepts

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

Reaction Manipulation
When dealing with equilibrium constants and reactions, sometimes it becomes necessary to manipulate given equations in order to derive a desired reaction. One common way this is done is by reversing the direction of a reaction or adding multiple reactions together. Each operation has implications for the equilibrium constant. For example, reversing a chemical reaction means you take the reciprocal of the equilibrium constant. Combining reactions involves multiplying their equilibrium constants. This approach allows you to express new reactions based on known ones, making problem-solving in chemistry more versatile.
Chemical Equilibrium
Chemical equilibrium refers to a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction. At equilibrium, the concentrations of reactants and products remain constant over time. This does not mean that the reactions have stopped, but rather that they occur with the same frequency in both directions. Equilibrium can be described by the equilibrium constant ( K ), which quantifies the concentrations of reactants and products at this state. A large equilibrium constant favors products, while a small one favors reactants. Understanding equilibrium is crucial for predicting and manipulating the outcomes of chemical reactions.
Thermodynamics
Thermodynamics involves the study of energy changes in chemical reactions. When an equilibrium constant is determined, it relates to the Gibbs free energy change ( 螖G ) of the reaction, which helps predict whether a reaction is spontaneous. The relationship between K and 螖G is given by the equation: 螖G = -RT lnK , where R is the gas constant and T is the temperature in Kelvin. If 螖G is negative, the reaction is spontaneous; if positive, it is non-spontaneous under the given conditions. Thus, thermodynamics provides insight into why certain reactions are more favorable than others and how energy is stored or released during the process.
Kp Calculations
Calculating the equilibrium constant in terms of pressure, K_p , is essential when dealing with gas reactions. K_p is similar to the concentration equilibrium constant K_c , but it uses the partial pressures of the gases involved. The relationship between K_c and K_p is given by K_p = K_c(RT)^{螖n} , where 螖n is the change in moles of gas. To determine K_p for a combination of reactions, as in the original exercise, it involves Reverse and add: taking reciprocal for reversed reaction equilibrium constant and multiplying for added reactions. This means the greater the understanding of K_p , the more you can predict outcomes of reactions involving gases in equilibrium accurately.

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Most popular questions from this chapter

At \(1200^{\circ} \mathrm{C}, K_{c}=2.5 \times 10^{4}\) for the equation $$ \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \leftrightharpoons 2 \mathrm{HCl}(g) $$ If \(0.50\) moles of \(\mathrm{H}_{2}(g)\) and \(0.50\) moles of \(\mathrm{Cl}_{2}(g)\) are introduced initially into a reaction vessel, how many moles of \(\mathrm{HCl}(g)\) are there at equilibrium?

Phosgene, \(\mathrm{COCl}_{2}(g)\), a toxic gas used in the synthesis of a variety of organic compounds, decomposes according to $$ \mathrm{COCl}_{2}(g) \leftrightharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ A sample of phosgene gas at an initial concentration of \(0.500 \mathrm{M}\) is heated at \(527^{\circ} \mathrm{C}\) in a reaction vessel. At equilibrium, the concentration of \(\mathrm{CO}(g)\) was found to be \(0.046 \mathrm{M}\). Calculate the equilibrium constant for the reaction equation at \(527^{\circ} \mathrm{C}\).

Nitrogen dioxide decomposes at high temperatures according to the equation $$ 2 \mathrm{NO}_{2}(g) \leftrightharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ Suppose initially we have pure \(\mathrm{NO}_{2}(g)\) at \(1000 \mathrm{~K}\) and \(0.500\) bar. If the total pressure is \(0.732\) bar when equilibrium is reached, what is the value of \(K_{\mathrm{p}}\) ? What are its units?

Given the equilibrium constants at \(1000 \mathrm{~K}\) for the following equations (1) \(\mathrm{CaCO}_{3}(s) \leftrightharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) \(K_{\mathrm{P}_{1}}=0.040 \mathrm{bar}\) (2) \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)\) \(K_{\mathrm{P}_{2}}=1.9 \mathrm{bar}\) determine the equilibrium constant, \(K_{\mathrm{p}}\), at \(1000 \mathrm{~K}\) for the equation (3) \(\mathrm{CaCO}_{3}(s)+\mathrm{C}(s) \leftrightharpoons \mathrm{CaO}(s)+2 \mathrm{CO}(g)\)

The value of the equilibrium constant for the chemical equation $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \leftrightharpoons 2 \mathrm{HI}(g) $$ is \(K_{\mathrm{p}}=85\) at \(280^{\circ} \mathrm{C}\). Calculate the value of \(K_{\mathrm{p}}\) at \(280^{\circ} \mathrm{C}\) for the equation $$ \mathrm{HI}(g) \leftrightharpoons \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{I}_{2}(g) $$
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