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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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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to the state in a chemical reaction when the concentrations of reactants and products remain constant over time. This happens because the forward and reverse reactions occur at the same rate. Consider it like a seesaw that's perfectly balanced, with neither side going up or down.
Once a reaction reaches equilibrium, no net change is observed in the concentrations of substances involved, even though at the molecular level, both reactions are still proceeding.
In our example of carbon monoxide reacting with hydrogen to form methane and water, equilibrium is reached when the rate at which CO and Hâ‚‚ react equals the rate at which CHâ‚„ and Hâ‚‚O break down back into CO and Hâ‚‚.
  • The equilibrium constant, K, is a numerical value that can provide insights into the concentrations of products and reactants at equilibrium.
  • A large value of K indicates that, at equilibrium, products are favored, whereas a small value of K suggests that reactants dominate.
Reaction Kinetics
Reaction kinetics deals with the rate of chemical reactions and what factors influence these rates. It's like understanding how fast or slow a reaction goes from start to finish.
Various factors affect the rate of reactions, such as temperature, pressure, concentration, and the presence of catalysts.
The reaction rate provides insight into how quickly reactants transform into products, which is critical for processes in both industrial and biological systems.
For the reaction of CO and Hâ‚‚ to CHâ‚„ and Hâ‚‚O, gaining insight into reaction kinetics helps us understand at what conditions the reaction happens fastest and most efficiently.
The temperature at which this specific reaction, at 1127°C, influences how rapidly equilibrium is achieved. Reaction kinetics provides tools to manipulate such reactions to one's advantage in industrial applications.
Law of Mass Action
The Law of Mass Action is a fundamental principle used to derive the equilibrium constant expression from a balanced chemical equation. It states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants, each raised to a power equal to its coefficient in the balanced equation.
This law forms the basis for calculating the equilibrium constant, K. By applying it, we can predict how the concentration of substances will shift as the reaction moves toward equilibrium.
In our case of CO reacting with Hâ‚‚, the law tells us to set up the expression for K as \(K = \frac{P_{CHâ‚„} \times P_{Hâ‚‚O}}{P_{CO} \times (P_{Hâ‚‚})^3}\).
  • This equation uses partial pressures because the reaction occurs in a gaseous state, and each substance's pressure is proportional to its concentration.
  • The exponents in the denominator indicate the reaction order for hydrogen, crucial for predicting how changes in pressure will affect equilibrium.
Understanding the Law of Mass Action empowers us to analyze how equilibrium will respond to changes in pressure, temperature, or concentration, thus making it an essential tool in chemical reactions.

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Most popular questions from this chapter

The system $$3 \mathrm{Z}(g)+\mathrm{Q}(g) \rightleftharpoons 2 \mathrm{R}(g)$$ is at equilibrium when the partial pressure of \(\mathrm{Q}\) is \(0.44 \mathrm{~atm} .\) Sufficient \(\mathrm{R}\) is added to increase the partial pressure of Q temporarily to \(1.5 \mathrm{~atm} .\) When equilibrium is reestablished, the partial pressure of \(Q\) could be which of the following? (a) \(1.5 \mathrm{~atm}\) (b) \(1.2 \mathrm{~atm}\) (c) \(0.80\) atm (d) \(0.44 \mathrm{~atm}\) (e) \(0.40 \mathrm{~atm}\)

A compound, \(\mathrm{X}\), decomposes at \(131^{\circ} \mathrm{C}\) according to the following equation: $$2 \mathrm{X}(g) \rightleftharpoons \mathrm{A}(g)+3 \mathrm{C}(g) \quad K=1.1 \times 10^{-3}$$ If a flask initially contains \(\mathrm{X}, \mathrm{A}\), and \(\mathrm{C}\), all at partial pressures of \(0.250 \mathrm{~atm}\), in which direction will the reaction proceed?

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 \(\operatorname{ICl}(g)\).

At \(460^{\circ} \mathrm{C}\), the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{NO}(g)+\mathrm{SO}_{3}(g)$$ has \(K=84.7\). All gases are at an initial pressure of \(1.25\) atm. (a) Calculate the partial pressure of each gas at equilibrium. (b) Compare the total pressure initially with the total pressure at equilibrium. Would that relation be true of all gaseous systems?

At a certain temperature, nitrogen and oxygen gases combine to form nitrogen oxide gas. $$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(g)$$ When equilibrium is established, the partial pressures of the gases are: \(P_{\mathrm{N}_{2}}=\) \(1.2 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.80 \mathrm{~atm}, P_{\mathrm{NO}}=0.022 \mathrm{~atm} .\) (a) Calculate \(K\) at the temperature of the reaction. (b) After equilibrium is reached, more oxygen is added to make its partial pressure \(1.2\) atm. Calculate the partial pressure of all gases when equilibrium is reestablished.

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