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Chapter 15: Problem 47

At \(1285^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is $K_{c}=1.04 \times 10^{-3} .\( A \)1.00-\mathrm{L}$ vessel containing an equilibrium mixture of the gases has \(1.50 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel?

Short Answer

Expert verified
The mass of Br(g) in the vessel at equilibrium is approximately 0.72 g.

Step by step solution

01

Write the balanced chemical equation and Kc expression

The balanced chemical equation for the reaction is already given: \[ \mathrm{Br_2(g)} \rightleftharpoons 2\mathrm{Br(g)} \] The Kc expression for this reaction is: \[ K_c = \frac{[\mathrm{Br}]^2}{[\mathrm{Br_2}]} \] Where [Br] and [Br2] represent the molar concentrations of the bromine atoms and bromine gas at equilibrium, respectively.
02

Convert mass of Br2 to moles

We need to convert the given mass of Br2 (1.50 g) into moles using its molar mass. The molar mass of Br2 is: \[ \mathrm{M(Br_2)} = 2 \times \mathrm{M(Br)} = 2 \times 79.904 = 159.808\,\mathrm{g/mol} \] Now, calculate the moles of Br2: \[ \mathrm{moles\,of\,Br_2} = \frac{\mathrm{mass\,of\,Br_2}}{\mathrm{molar\,mass\,of\,Br_2}} = \frac{1.50\,\mathrm{g}}{159.808\,\mathrm{g/mol}} =9.39\,\mathrm{x\,10^{-3}\,mol} \]
03

Set up an ICE table

We will set up an ICE (Initial, Change, Equilibrium) table to find the molar concentrations of Br2 and Br at equilibrium. | | Br2 | -> | 2 Br | |-------|-----|------|-------| |Initial| x | | 0 | |Change |-y | | 2y | |Equilibrium| x-y | | 2y | Where x represents the initial moles of Br2 and y represents moles of Br2 that dissociate at equilibrium. We already calculated x as 9.39 x 10^-3 mol.
04

Calculate the molar concentrations and solve for y

Now, we will use the Kc expression and the ICE table to solve for y. \[ K_c = \frac{[\mathrm{Br}]^2}{[\mathrm{Br_2}]} = \frac{(2\,y)^2}{x - y} \] The vessel has a volume of 1.00 L, so the molar concentrations at equilibrium can be calculated like this: \[ 1.04 \times 10^{-3} = \frac{((2\,y)/1.00)^2}{(x - y)/1.00} \] Now, solve for y: \[ y = 0.0045 \,\text{mol} \]
05

Calculate the mass of Br at equilibrium

Since two moles of Br are produced for every mole of Br2 dissociated, the number of moles of Br at equilibrium is twice the value of y: \[ \mathrm{moles\,of\,Br} = 2 \times y = 2 \times 0.0045 \,\text{mol} = 0.0090\,\text{mol} \] Now, we can calculate the mass of Br using its molar mass: \[ \mathrm{mass\,of\,Br} = \mathrm{moles\,of\,Br} \times \mathrm{M(Br)} = 0.0090 \,\text{mol} \times 79.904 \,\text{g/mol} \approx 0.72\,\text{g} \] Thus, there is approximately 0.72 g of Br(g) in the vessel at equilibrium.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\), the reaction $$\mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q)$$ has an equilibrium constant \(K_{c}=7.1 \times 10^{-4}\). What are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and $\mathrm{CrO}_{4}{ }^{2-}\( in a saturated solution of \)\mathrm{CaCrO}_{4} ?$

If \(K_{c}=0.013 \mathrm{~L} / \mathrm{mol}\) for $2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$ at \(1000 \mathrm{~K}\), what is the value of \(K_{p}\) for this reaction at this temperature?

Gaseous hydrogen iodide is placed in a closed container at $450^{\circ} \mathrm{C},\( where it partially decomposes to hydrogen and iodine: \)2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) .$ At equilibrium it is found that $[\mathrm{HI}]=4.50 \times 10^{3} \mathrm{M},\left[\mathrm{H}_{2}\right]=5.75 \times 10^{4} \mathrm{M}$, and \(\left[\mathrm{I}_{2}\right]=5.75 \times 10^{-4} \mathrm{M}\). What is the value of \(K_{c}\) at this temperature?

The equilibrium constant for the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ is \(K_{c}=1.3 \times 10^{-2}\) at \(1000 \mathrm{~K}\). (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr? (b) Calculate \(K_{c}\) for $2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) .$ (c) Calculate \(K_{c}\) for $\mathrm{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)$.

Bromine and hydrogen react in the gas phase to form hydrogen bromide: \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) .\) The reaction enthalpy is \(\Delta H^{\circ}=-6 \mathrm{~kJ} .\) (a) To increase the equilibrium yield of hydrogen bromide would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen bromide by controlling the pressure of this reaction? If so, would high or low pressure favor formation of \(\mathrm{HBr}(g) ?\)
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