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Chapter 15: Problem 75

Starting with \(\mathrm{SO}_{3}(\mathrm{g})\) at \(1.00 \mathrm{atm},\) what will be the total pressure when equilibrium is reached in the following reaction at \(700 \mathrm{K} ?\) \(2 \mathrm{SO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.6 \times 10^{-5}\)

Short Answer

Expert verified
The total equilibrium pressure of the system at \(700 \mathrm{K}\) is 0.9984 atm.

Step by step solution

01

Define Initial Conditions and Reactions

Initially, there is 1 atm of \(SO_{3}\) and no \(SO_{2}\) or \(O_{2}\). When the reaction reaches equilibrium, some \(SO_{3}\) will have converted into \(SO_{2}\) and \(O_{2}\). Define \(x\) to be the change in pressure of \(SO_{3}\), so the change in \(SO_{3}\) is -2x (since it is being consumed), and the changes in \(SO_{2}\) and \(O_{2}\) are +2x and +x, respectively (since they are being produced).
02

Formulate the ICE Table

Write down the initial, change, and equilibrium expressions for all species involved: initial pressures are \(SO_{3}: 1, SO_{2}: 0, O_{2}: 0\), changes in pressures are \(SO_{3}: -2x, SO_{2}: +2x, O_{2}: +x\), and equilibrium pressures are \(SO_{3}: 1-2x, SO_{2}: 2x, O_{2}: x\).
03

Formulate the Equilibrium Constant Expression

Next, write down the equilibrium constant \(K_{\mathrm{p}}\) expression based on the balanced chemical equation: \(K_{\mathrm{p}} = \frac{{[SO_{2}]^2[O_{2}]}}{{[SO_{3}]^2}}\) and substitute the equilibrium pressures from the ICE table, yielding \(1.6 \times 10^{-5} = \frac{{(2x)^2 \cdot x}}{{(1-2x)^2}}\).
04

Solve for x

The equation in Step 3 yields a cubic equation, which can be challenging to solve algebraically. However, making an assumption that x << 1 (since \(K_{\mathrm{p}}\) is very small indicating that not much reactant is converted to product), the equation simplifies to: \(1.6 \times 10^{-5} = 4x^3\), which can be solved to find x = 0.0012.
05

Calculate Final Pressure at Equilibrium

Substitute x = 0.0012 into the expressions for total pressure: \(P_{total} = P_{SO_{3}} + P_{SO_{2}} + P_{O_{2}} = (1 - 2 \cdot 0.0012) + 2 \cdot 0.0012 + 0.0012 = 0.9984 \: atm\).

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Most popular questions from this chapter

The following reaction is used in some self-contained breathing devices as a source of \(\mathrm{O}_{2}(\mathrm{g})\) $$\begin{aligned} 4 \mathrm{KO}_{2}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s}) &+3 \mathrm{O}_{2}(\mathrm{g}) \\\ K_{\mathrm{p}} &=28.5 \mathrm{at} 25^{\circ} \mathrm{C} \end{aligned}$$ Suppose that a sample of \(\mathrm{CO}_{2}(\mathrm{g})\) is added to an evacuated flask containing \(\mathrm{KO}_{2}(\mathrm{s})\) and equilibrium is established. If the equilibrium partial pressure of \(\mathrm{CO}_{2}(\mathrm{g})\) is found to be \(0.0721 \mathrm{atm},\) what are the equilibrium partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) and the total gas pressure?

One of the key reactions in the gasification of coal is the methanation reaction, in which methane is produced from synthesis gas-a mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons & \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ \Delta H &=-230 \mathrm{kJ} ; K_{\mathrm{c}}=190 \mathrm{at} 1000 \mathrm{K} \end{aligned}$$ (a) Is the equilibrium conversion of synthesis gas to methane favored at higher or lower temperatures? Higher or lower pressures? (b) Assume you have 4.00 mol of synthesis gas with a 3:1 mol ratio of \(\mathrm{H}_{2}(\mathrm{g})\) to \(\mathrm{CO}(\mathrm{g})\) in a 15.0 L flask. What will be the mole fraction of \(\mathrm{CH}_{4}(\mathrm{g})\) at equilibrium at \(1000 \mathrm{K} ?\)

The volume of the reaction vessel containing an equilibrium mixture in the reaction \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) is increased. When equilibrium is re-established, (a) the amount of \(\mathrm{Cl}_{2}\) will have increased; (b) the amount of \(\mathrm{SO}_{2}\) will have decreased; (c) the amounts of \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) will have remained the same; (d) the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) will have increased.

The standard enthalpy of reaction for the decomposition of calcium carbonate is \(\Delta H^{\circ}=813.5 \mathrm{kJmol}^{-1}\) As temperature increases, does the concentration of calcium carbonate increase, decrease, or remain the same? Explain.

Would you expect that the amount of \(\mathrm{N}_{2}\) to increase, decrease, or remain the same in a scuba diver's body as he or she descends below the water surface?
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