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Chapter 15: Problem 36

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 \(K_{c}\)

Short Answer

Expert verified
The equilibrium concentrations are [H₂] = 0.142 M, [Br₂] = 0.018 M, and [HBr] = 0.404 M. The equilibrium constant Kc is 639.

Step by step solution

01

Calculate the initial moles of reactants and moles of H₂ at equilibrium

First, we need to convert the given masses of H₂, Br₂, and the mass of H₂ at equilibrium into moles. We can use the molar masses of H₂ and Br₂ to do this: H₂ molar mass: 2g/mol Br₂ molar mass: 2 * 79.9g/mol = 159.8g/mol Initial moles of H₂: \(1.374 g \mathrm{H}_2 \times \frac{1 \mathrm{mol} \mathrm{H}_2}{2g} = 0.687 \mathrm{mol}\) Initial moles of Br₂: \(70.31 g \mathrm{Br}_2 \times \frac{1 \mathrm{mol} \mathrm{Br}_2}{159.8g} = 0.440 \mathrm{mol}\) Moles of H₂ at equilibrium: \(0.566 g \mathrm{H}_2 \times \frac{1 \mathrm{mol} \mathrm{H}_2}{2g} = 0.283 \mathrm{mol}\)
02

Calculate moles of reactants and products at equilibrium

Since we know the moles of H₂ decreased by (0.687 - 0.283) = 0.404 mol, we can deduce the loss of moles of Br₂ and the gain in moles of HBr: Moles of Br₂ at equilibrium = 0.440 - 0.404 = 0.036 mol (as 1 mol of Br₂ is used per mol of H₂ used) Moles of HBr at equilibrium = 2 * 0.404 = 0.808 mol (as 2 mol of HBr is formed per mol of H₂ used)
03

Calculate the equilibrium concentrations

We can now find the equilibrium concentrations of each substance by dividing the moles at equilibrium by the volume of the vessel: Concentration of H₂: \(\frac{0.283 \mathrm{mol}}{2.00 L} = 0.142 M\) Concentration of Br₂: \(\frac{0.036 \mathrm{mol}}{2.00 L} = 0.018 M\) Concentration of HBr: \(\frac{0.808 \mathrm{mol}}{2.00 L} = 0.404 M\) Thus, the equilibrium concentrations are: [H₂] = 0.142 M, [Br₂] = 0.018 M, and [HBr] = 0.404 M.
04

Calculate the equilibrium constant Kc

Using the balanced equation, we can write the expression for Kc: \(K_c = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_2][\mathrm{Br}_2]}\) Plugging in the equilibrium concentrations, we find Kc: \(K_c = \frac{(0.404 M)^2}{(0.142 M)(0.018 M)} = 639\) The equilibrium constant Kc for this reaction at 700 K is 639.

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Most popular questions from this chapter

Which of the following statements are true and which are false? (a) For the reaction $2 \mathrm{~A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{A}_{2} \mathrm{~B}(g) K_{c}$ and \(K_{p}\) are numerically the same. (b) It is possible to distinguish \(K_{c}\) from \(K_{p}\) by comparing the units used to express the equilibrium constant. \((\mathbf{c})\) For the equilibrium in (a), the value of \(K_{c}\) increases with increasing pressure.

At \(120^{\circ} \mathrm{C}, K_{c}=0.090\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ In an equilibrium mixture of the three gases, the concentrations of \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2}\) are $0.100 \mathrm{M}\( and \)0.075 \mathrm{M}$, respectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15 ), \(K_{p}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ is about \(1 \times 10^{-11}\). Assuming that the exhaust gas (total pressure \(101.3 \mathrm{kPa}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2},\) and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\mathrm{mol} \%\).

A mixture of \(0.140 \mathrm{~mol}\) of \(\mathrm{NO}, 0.060 \mathrm{~mol}\) of \(\mathrm{H}_{2},\) and 0.260 mol of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a \(2.0-\mathrm{L}\) vessel at \(330 \mathrm{~K}\). Assume that the following equilibrium is established: $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ At equilibrium \(\left[\mathrm{H}_{2}\right]=0.010 \mathrm{M}\). (a) Calculate the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2},\) and \(\mathrm{H}_{2} \mathrm{O} .\) (b) Calculate \(K_{c}\).

As shown in Table \(15.2,\) the equilibrium constant for the reaction $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\( is \)K_{p}=4.23 \times 10^{-7}\( at \)300^{\circ} \mathrm{C}\(. Pure \)\mathrm{NH}_{3}$ is placed in a 1.00-L flask and allowed to reach equilibrium at this temperature. There are $1.05 \mathrm{~g} \mathrm{NH}_{3}$ in the equilibrium mixture. (a) What are the masses of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in the equilibrium mixture? (b) What was the initial mass of ammonia placed in the vessel? (c) What is the total pressure in the vessel?
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