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Chapter 14: Problem 85

A quantity of 0.20 mole of carbon dioxide was heated to a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ Under these conditions, the average molar mass of the gases was \(35 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). (b) What is \(K_{P}\) if the total pressure is 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.

Short Answer

Expert verified
Calculate the moles of \(CO_{2}\) by assuming 'x' moles react. Use the given formula for average molar mass to solve for x. Then calculate the mole fractions of \(CO_{2}\) and \(CO\), and use these to calculate \(K_{P}\). Substitute these values into the formula for a numerical solution.

Step by step solution

01

Understand the reaction and mole fractions definition

The equilibrium reaction is given by: \(C(s)+CO_{2}(g) \Leftrightarrow 2 CO(g)\). In order to calculate the mole fractions, we need to know how many moles of each gas are present at equilibrium.
02

Calculate the moles of \(CO_{2}\)

We are given that 0.20 mole of \(CO_{2}\) was heated. Let's assume that 'x' moles of \(CO_{2}\) react, thus producing '2x' moles of \(CO\). So, the moles of \(CO_{2}\) left at equilibrium will be 0.20 - x, and the moles of \(CO\) will be 2x.
03

Calculate the average molar mass and solve for x

We know that the average molar mass is given by the sum of the products of the molar mass of each gas and its respective mole fraction. We therefore have: \[ 35 = (44 \times \frac{(0.20 - x)}{(0.20 - x + 2x)}) + (28 \times \frac{2x}{(0.20 - x + 2x)})\]Solving this for x will give the moles of \(CO_{2}\) and \(CO\)
04

Calculate mole fractions

Now that we have the value of x, we can substitute it back into the fraction formulas. The mole fraction of \(CO_{2}\) is by definition: \(\frac{(0.20 - x)}{(0.20 - x + 2x)}\)and the mole fraction of \(CO\) is: \(\frac{2x}{(0.20 - x + 2x)}\) . These will give us the respective mole fractions.
05

Calculate the value of \(K_{P}\)

From the law of mass action, we know that \(K_{P}\) is given by the ratio of the partial pressures of the products to that of the reactants, each raised to their respective stoichiometric coefficients. Given that the total pressure is 11 atm, we find partial pressures of \(CO_{2}\) and \(CO\), using the mole fractions and the total pressure. We can then substitute these into the expression for \(K_{P}\): \[K_{P} = \frac{(PCO)^{2}}{PCO_{2}}\]
06

Substitute the values

Substitute the numerical values of the partial pressures into the formula to find the numerical value of \(K_{P}\).

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Most popular questions from this chapter

The decomposition of ammonium hydrogen sulfide $$\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ is an endothermic process. A 6.1589 -g sample of the solid is placed in an evacuated \(4.000-\mathrm{L}\) vessel at exactly \(24^{\circ} \mathrm{C}\). After equilibrium has been established, the total pressure inside is 0.709 atm. Some solid \(\mathrm{NH}_{4} \mathrm{HS}\) remains in the vessel. (a) What is the \(K_{P}\) for the reaction? (b) What percentage of the solid has decomposed? (c) If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel?

Outline the steps for calculating the concentrations of reacting species in an equilibrium reaction.

One mole of \(\mathrm{N}_{2}\) and three moles of \(\mathrm{H}_{2}\) are placed in a flask at \(375^{\circ} \mathrm{C}\). Calculate the total pressure of the system at equilibrium if the mole fraction of \(\mathrm{NH}_{3}\) is \(0.21 .\) The \(K_{P}\) for the reaction is \(4.31 \times 10^{-4}\).

Determine the initial and equilibrium concentrations of HI if the initial concentrations of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) are both \(0.16 M\) and their equilibrium concentrations are both \(0.072 M\) at \(430^{\circ} \mathrm{C}\). The equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) is 54.2 at \(430^{\circ} \mathrm{C}\)

Briefly describe the importance of equilibrium in the study of chemical reactions.
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