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The equilibrium constant for the chemical equation $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g) $$ at \(1000 \mathrm{~K}\) is \(1.90\) bar. If the equilibrium pressure of \(\mathrm{CO}(g)\) is \(1.50\) bar, what is the equilibrium pressure of \(\mathrm{CO}_{2}(g) ?\)

Short Answer

Expert verified
The equilibrium pressure of \(\mathrm{CO}_2\) is approximately 1.18 bar.

Step by step solution

01

Organize Given Information

We are provided with the chemical reaction: \( \mathrm{C}(s) + \mathrm{CO}_2(g) \leftrightharpoons 2 \mathrm{CO}(g) \). The given equilibrium constant \( K_p = 1.90 \) bar at \( 1000\, \mathrm{K} \). The equilibrium pressure of \( \mathrm{CO}(g) = 1.50 \) bar.
02

Write the Expression for Equilibrium Constant

The expression for the equilibrium constant \( K_p \) for the given reaction is: \[ K_p = \frac{{P_{\mathrm{CO}}^2}}{{P_{\mathrm{CO}_2}}} \] where \( P_{\mathrm{CO}} \) and \( P_{\mathrm{CO}_2} \) are the equilibrium pressures of \( \mathrm{CO} \) and \( \mathrm{CO}_2 \) respectively.
03

Substitute Known Values

Substitute the known values into the equilibrium expression: \[ 1.90 = \frac{{(1.50)^2}}{{P_{\mathrm{CO}_2}}} \].
04

Solve for Unknown Pressure

Rearrange the equation to solve for \( P_{\mathrm{CO}_2} \):\[ P_{\mathrm{CO}_2} = \frac{{(1.50)^2}}{1.90} \].
05

Calculate the Equilibrium Pressure of \(\mathrm{CO}_2\)

Calculate the value: \[ P_{\mathrm{CO}_2} = \frac{{2.25}}{1.90} \approx 1.184 \text{ bar} \]. Therefore, the equilibrium pressure of \( \mathrm{CO}_2(g) \) is approximately \( 1.18 \) bar.

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Key Concepts

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

Equilibrium Constant
In chemical reactions, the equilibrium constant (\( K_p \)) is a crucial figure. It links the concentrations or pressures of reactants and products at equilibrium. For gas-phase reactions like this, the expression utilizes partial pressures. Each pressure is raised to the power of its coefficient in the balanced equation. Consider the equation: \[\mathrm{C}(s) + \mathrm{CO}_2(g) \leftrightharpoons 2 \mathrm{CO}(g)\]While solids like carbon (\( \mathrm{C} \)) do not appear in the\( K_p \) calculation, the gas pressures are included. This is a product over reactant format:\[K_p = \frac{{\left( P_{\mathrm{CO}} \right)^2}}{{P_{\mathrm{CO}_2}}}\]So, knowing\( K_p \) helps predict how much product and reactant will form at equilibrium.
  • If\( K_p \) is large, products dominate.
  • If\( K_p \) is small, reactants prevail.
This allows chemists to understand and influence reactions.
Partial Pressure
Partial pressure refers to the individual pressure exerted by a single gas component in a mixture. In equilibrium calculations, especially in gaseous systems like with \(\mathrm{CO}_2(g)\) and \(\mathrm{CO}(g)\), partial pressure is vital. It directly affects the equilibrium state of the reaction.
Consequently, if the total pressure of the system changes, the equilibrium condition shifts. The partial pressure of each gas depends on:
  • The total pressure of the system.
  • The mole fraction of the gas in the mixture.
In this exercise, \( P_{\mathrm{CO}} \) is given as\(1.50 \text{ bar} \), and by rearranging the equilibrium constant equation:\[P_{\mathrm{CO}_2} = \frac{{\left(1.50\right)^2}}{1.90} \approx 1.18 \text{ bar}\]This shows how the partial pressures govern the equilibrium and balance of the chemical composition.
Le Chatelier's Principle
Le Chatelier's Principle is a guiding tool for predicting the effect of changes on a chemical equilibrium. It asserts that if a dynamic equilibrium is disturbed by changing the conditions, the system adjusts to counteract the imposed change.
In practical terms, this principle can be applied by observing pressure shifts in gas reactions. For instance:
  • Increasing the total pressure in a reaction involving gases generally leads to a shift towards the side with fewer gas molecules.
  • Decreasing the pressure tends to shift the equilibrium towards the side with more gas molecules.
Using Le Chatelier's insight, if the external conditions of the reaction \(\mathrm{C}(s) + \mathrm{CO}_2(g) \leftrightharpoons 2 \mathrm{CO}(g) \) change, such as by adding or removing \(\mathrm{CO}_2(g)\), the system will adjust to maintain equilibrium.
Understanding this principle allows chemists to control reactions more effectively, optimizing yields and minimizing unwanted by-products in industrial processes.

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Most popular questions from this chapter

The equilibrium constant at \(823 \mathrm{~K}\) for the chemical equation $$ \mathrm{MgCl}_{2}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \leftrightharpoons \mathrm{Mg} \mathrm{O}(s)+\mathrm{Cl}_{2}(g) $$ is \(K_{\mathrm{p}}=1.75 \mathrm{~atm}^{1 / 2}\). Suppose that \(50.0\) grams of \(\mathrm{MgCl}_{2}(s)\) is placed in a reaction vessel with \(2.00\) liters of oxygen at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) and that the reaction vessel is sealed and heated to \(823 \mathrm{~K}\) until equilibrium is attained. Calculate \(P_{\mathrm{Cl}_{2}}\) and \(P_{\mathrm{O}_{2}}\) at equilibrium in atm.

Hydrogen sulfide decomposes at \(1400 \mathrm{~K}\) according to the chemical equation $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \leftrightharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ Suppose initially we have pure \(\mathrm{H}_{2} \mathrm{~S}(g)\) at a pressure of \(0.956\) bar. If the total pressure is \(1.26\) bar when equilibrium is reached, what is the value of \(K_{\mathrm{p}}\) and its corresponding 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 decomposition of ammonium carbamate, \(\mathrm{NH}_{2} \mathrm{COONH}_{4}(s)\), takes place according to the chemical equation $$ \mathrm{NH}_{2} \mathrm{COONH}_{4}(s) \leftrightharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ Show that if all the \(\mathrm{NH}_{3}(g)\) and \(\mathrm{CO}_{2}(g)\) result from the decomposition of ammonium carbamate, then \(K_{\mathrm{p}}=\) \((4 / 27) P^{5}\), where \(P\) is the total pressure at equilibrium.

Tin can be prepared by heating \(\mathrm{SnO}_{2}(s)\) ore with hydrogen gas, according to: $$ \mathrm{SnO}_{2}(s)+2 \mathrm{H}_{2}(g) \leftrightharpoons \mathrm{Sn}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ When the reactants are heated to \(500^{\circ} \mathrm{C}\) in a closed vessel, \(\left[\mathrm{H}_{2} \mathrm{O}\right]=\left[\mathrm{H}_{2}\right]=0.25 \mathrm{M}\) at equilibrium. If more hydrogen is added so that its new initial concentration is \(0.50 \mathrm{M}\), what are the concentrations of \(\mathrm{H}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) when equilibrium is restored?

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