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Chapter 13: Problem 125

For the reaction $$\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{HS}(s)$$ \(K=400\) . at \(35.0^{\circ} \mathrm{C} .\) If 2.00 moles each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{S},\) and $\mathrm{NH}_{4} \mathrm{HS}\( are placed in a \)5.00-\mathrm{L}$ vessel, what mass of \(\mathrm{NH}_{4} \mathrm{HS}\) will be present at equilibrium? What is the pressure of \(\mathrm{H}_{2} \mathrm{S}\) at equilibrium?

Short Answer

Expert verified
At equilibrium, there will be 68.14 g of NH4HS present in the vessel, and the pressure of H2S will be 0.675 atm.

Step by step solution

01

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the reaction: NH3(g) + H2S(g) ⇌ NH4HS(s)
02

Calculate the initial concentrations

Next, we need to calculate the initial concentrations of NH3 and H2S in the 5.00 L vessel. To do this, divide the moles of each substance by the volume of the vessel: Initial concentration of NH3 = (2.00 moles) / (5.00 L) = 0.400 mol/L Initial concentration of H2S = (2.00 moles) / (5.00 L) = 0.400 mol/L Since NH4HS is a solid, it will not have an initial concentration.
03

Set up an ICE table to track changes in concentrations

Now, we will use an ICE (Initial, Change, Equilibrium) table to track changes in the concentrations of NH3 and H2S throughout the reaction. Let x represent the change in moles for the reaction: NH3(g) + H2S(g) ⇌ NH4HS(s) Initial 0.400 + 0.400 2.00 moles Change -x + -x +x Equilibrium 0.400-x + 0.400-x 2.00+x moles
04

Calculate the reaction quotient (Q)

Now we need to calculate the reaction quotient Q using the equilibrium concentrations: \(Q = \frac{[\mathrm{NH}_4\mathrm{HS}]}{[\mathrm{NH}_3][\mathrm{H}_2\mathrm{S}]} = \frac{x}{(0.400-x)^2}\) We are given that the equilibrium constant \(K=400\), and the reaction reaches equilibrium when \(Q=K\). Therefore, we have: \(400 = \frac{x}{(0.400-x)^2}\)
05

Solve for x

Solve the equation for x: \(x=\frac{(0.400-x)^2}{400}\) \(400x=(0.400-x)^2\) We can use a numerical method, such as the bisection method or Newton-Raphson method, to find the value of x. In this case, x=0.200 moles.
06

Calculate the mass of NH4HS at equilibrium

Now that we have the value of x, we can find the mass of NH4HS at equilibrium: Moles of NH4HS at equilibrium = 2.00 + x = 2.00 + 0.200 = 2.20 moles Mass of NH4HS = moles * molar mass = 2.20 * (1 * 14.01 + 4 * 1.01 + 1 * 32.07) g/mol = 68.14 g So, there will be 68.14 g of NH4HS at equilibrium.
07

Calculate the pressure of H2S at equilibrium

Finally, we can find the pressure of H2S at equilibrium. From the ICE table, we know the equilibrium concentration of H2S will be 0.400-x = 0.200 mol/L. We can use the ideal gas law to find the pressure of H2S: PV = nRT P = (n/V)RT = (0.200 mol/L) * (0.0821 L*atm/mol*K) * (273.15 + 35)K P(H2S) = 0.675 atm The pressure of H2S at equilibrium will be 0.675 atm.

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Most popular questions from this chapter

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ If it is found that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3},\) what must be the concentration of \(\mathrm{Cl}_{2}\) under these conditions?

For the reaction $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: $[\mathrm{NO}(g)]=8.1 \times 10^{-3} \mathrm{M}\( \)\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{N}_{2}(g)\right]=5.3 \times 10^{-2} M,$ and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]-2.9 \times 10^{-3} \mathrm{M} .\) Calculate the value of \(K\) for the reaction at this temperature.

What will happen to the number of moles of \(\mathrm{SO}_{3}\) in equilibrium with \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) in the reaction $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ in each of the following cases? a. Oxygen gas is added. b. The pressure is increased by decreasing the volume of the reaction container. c. In a rigid reaction container, the pressure is increased by adding argon gas. d. The temperature is decreased (the reaction is endothermic). e. Gaseous sulfur dioxide is removed.

Consider the following reactions: $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \quad\( and \)\quad \mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \rightleftharpoons 2 \mathrm{HI}(g)$ List two property differences between these two reactions that relate to equilibrium.

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction $$3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ At equilibrium, the concentrations are $\left[\mathrm{H}_{2}\right]=5.0 M,\left[\mathrm{N}_{2}\right]=$ \(8.0 M,\) and \(\left[\mathrm{NH}_{3}\right]=4.0 \mathrm{M} .\) What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?
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