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Chapter 15: Problem 81

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}\( taken together) is \)62.21 \mathrm{kPa}\(. What is \)K_{p}$ for this equilibrium at \(24^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The value of Kp for the given equilibrium reaction at 24°C is approximately 967.980 kPa².

Step by step solution

01

We start with an evacuated flask containing only NH4SH solid. There are no gases present initially, so the initial partial pressures of NH3 and H2S are both zero. #Step 2: Determine the change in the partial pressure of gases#

As the reaction proceeds, NH4SH solid decomposes and an equal amount of NH3 and H2S gas are formed. Let the change in partial pressure of NH3 and H2S be 'x'. At equilibrium, the partial pressure of NH3 is x, and the partial pressure of H2S is also x. #Step 3: Write the expression for Kp#
02

For the given reaction: NH4SH(s) NH3(g) + H2S(g) The expression for Kp at equilibrium is given by: Kp \(= \frac{P_{NH3} \times P_{H2S}}{P_{NH4SH}} \) Since NH4SH is a solid, its partial pressure does not affect the Kp value. Therefore, the expression becomes: Kp \(= P_{NH3} \times P_{H2S} \) #Step 4: Total pressure and partial pressures of gases at equilibrium#

At equilibrium, the total pressure is given as: Total pressure = Partial pressure of NH3 + Partial pressure of H2S We're given that the total pressure at equilibrium is 62.21 kPa. Since we know that at equilibrium both NH3 and H2S have the same partial pressure (x) we can write: 62.21 kPa = x + x #Step 5: Solve for partial pressure 'x'#
03

Combining the terms for x, we get: 62.21 kPa = 2x Solve for x: x = 31.105 kPa So, the partial pressure of NH3 at equilibrium is 31.105 kPa, and the partial pressure of H2S is also 31.105 kPa. #Step 6: Calculate Kp#

Now, we can use the partial pressures obtained in step 5 to find the Kp value: Kp \(= P_{NH3} \times P_{H2S} \) Kp = (31.105 kPa) × (31.105 kPa) Kp = 967.980 kPa² The value of Kp for the given equilibrium reaction at 24°C is approximately 967.980 kPa².

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Most popular questions from this chapter

(a) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

Consider the following equilibrium: $2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) \quad K_{c}=1.08 \times 10^{7}\( at \)700^{\circ} \mathrm{C}$ (a) Calculate \(K_{p \cdot}\) (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly $\mathrm{H}_{2} \mathrm{~S} ?(\mathbf{c})\( Calculatethevalue of \)K_{c}$ if you rewrote the equation $\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{~S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$

True or false: When the temperature of an exothermic reaction increases, the rate constant of the forward reaction decreases, which leads to a decrease in the equilibrium constant, \(K_{c}\)

Consider the following equilibrium between oxides of nitrogen $$3 \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g)$$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?

A mixture of \(1.374 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and \(70.31 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) is heated in a 2.00-L vessel at \(700 \mathrm{~K}\). These substances react according to $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ At equilibrium, the vessel is found to contain \(0.566 \mathrm{~g}\) of \(\mathrm{H}_{2}\). (a) Calculate the equilibrium concentrations of $\mathrm{H}_{2}, \mathrm{Br}_{2},\( and \)\mathrm{HBr} .$ (b) Calculate \(K_{c}\)
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