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Chapter 15: Problem 98

In Section 11.5 we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor and the corresponding expression for \(K_{p} .\) (b) By using data in Appendix \(\mathrm{B}\), give the value of \(K_{p}\) for this reaction at \(30^{\circ} \mathrm{C} .(\mathrm{c})\) What is the value of \(K_{p}\) for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

Short Answer

Expert verified
The equilibrium between liquid water and water vapor is represented by the equation \(H_{2}O(l) \rightleftharpoons H_{2}O(g)\). The expression for the equilibrium constant, Kp, is \(K_{p} = P(H_{2}O(g))\). Using data from Appendix B, the value of Kp for this reaction at 30 degrees Celsius is \(K_{p} = 4.24 \: kPa\). The value of Kp for any liquid in equilibrium with its vapor at the normal boiling point of the liquid is \(K_{p} = 101.3 \: kPa\).

Step by step solution

01

Write the Equation for the Equilibrium

Liquid water and water vapor are in equilibrium as per the following reaction: \[H_{2}O(l) \rightleftharpoons H_{2}O(g)\]
02

Write the Expression for the Equilibrium Constant (Kp)

The equilibrium constant Kp is expressed in terms of the partial pressures of the components in the reaction. In this case, the liquid water has no contribution to Kp since it has no partial pressure. Therefore, the expression for Kp becomes: \[K_{p} = \frac{P(H_{2}O(g))}{1}\] which simplifies to: \[K_{p} = P(H_{2}O(g))\]
03

Find the Value of Kp at 30 Degrees Celsius

To find the value of Kp at 30 degrees Celsius, we need the vapor pressure of water at this temperature. Using data from Appendix B, we find that: \[P(H_{2}O(g)) = 4.24 \: kPa\] Therefore, using the Kp expression from Step 2, we get: \[K_{p} = 4.24 \: kPa\]
04

Determine the Value of Kp at the Normal Boiling Point

At the normal boiling point of a liquid, the vapor pressure is equal to 1 atm or 101.3 kPa. Since Kp is equal to the vapor pressure of the substance in equilibrium with its liquid, the value of Kp for any liquid at its normal boiling point is: \[K_{p} = 101.3 \: kPa\]

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Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2}\) : $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ (a) Use thermochemical data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}\) for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g)+\mathrm{Cl}(g) $$ At \(25^{\circ} \mathrm{C}\) the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (b) Are reactants or products more plentiful at equilibrium? (c) What additional information would you need in order to decide whether the reaction as written is endothermic or exothermic?

Solid \(\mathrm{NH}_{4} \mathrm{SH}\) is introduced into an evacuated flask at \(24{ }^{\circ} \mathrm{C}\). The following reaction takes place: $$\mathrm{NH}_{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the total pressure (for \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) taken together) is 0.614 atm. What is \(K_{p}\) for this equilibrium at \(24^{\circ} \mathrm{C}\) ?

Suppose that the gas-phase reactions \(\mathrm{A} \longrightarrow \mathrm{B}\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) are both elementary processes with rate constants of \(4.7 \times 10^{-3} \mathrm{~s}^{-1}\) and \(5.8 \times 10^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value of the equilibrium constant for the equilibrium \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?\) (b) Which is greater at equilibrium, the partial pressure of \(\mathrm{A}\) or the partial pressure of \(\mathrm{B}\) ? Explain.

Can the equilibrium constant ever be zero? Explain.
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