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Chapter 15: Problem 35

Starting with \(0.3500 \mathrm{mol} \mathrm{CO}(\mathrm{g})\) and \(0.05500 \mathrm{mol}\) \(\mathrm{COCl}_{2}(\mathrm{g})\) in a \(3.050 \mathrm{L}\) flask at \(668 \mathrm{K},\) how many moles of \(\mathrm{Cl}_{2}(\mathrm{g})\) will be present at equilibrium? $$\begin{aligned} \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{g}) & \\ K_{\mathrm{c}} &=1.2 \times 10^{3} \mathrm{at} \ 668 \mathrm{K} \end{aligned}$$

Short Answer

Expert verified
After solving the quadratic equation, one would find the value of \(x\) which can be substituted into the equilibrium concentrations. The value of \(x\), representing the moles of \(Cl_2\) at equilibrium, is the required answer.

Step by step solution

01

- Write down the balanced chemical equation

The balanced chemical equation is: \(\mathrm{CO}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{COCl}_{2}\)
02

- Set up the initial and equilibrium concentrations

First, find the initial concentrations of \(\mathrm{CO}\) and \(\mathrm{COCl}_{2}\) by dividing the number of mols by the volume of the flask. The initial concentration of \(\mathrm{COCl}_{2}\) is \(0.05500 \mathrm{mol} / 3.050 \mathrm{L} = 0.0180 \mathrm{M}\) and the initial concentration of \(\mathrm{CO}\) is \(0.3500 \mathrm{mol} / 3.050 \mathrm{L} = 0.115 \mathrm{M}\). At equilibrium, the concentration of \(\mathrm{Cl}_{2}\) and \(\mathrm{CO}\) decrease by x and the concentration of \(\mathrm{COCl}_{2}\) increases by x. Therefore, the equilibrium concentrations are: \(\mathrm{CO}: 0.115 - x\), \(\mathrm{Cl}_{2}: x\), and \(\mathrm{COCl}_{2}: 0.0180 + x\).
03

- Set up the equilibrium expression and solve for x

The equilibrium expression for the reaction is: \(K_{\mathrm{c}} = [\mathrm{COCl}_{2}]/([\mathrm{CO}][\mathrm{Cl}_{2}])\). Substitute the equilibrium concentrations and the \(K_{\mathrm{c}}\)-value into this expression and solve for x: \(1.2 \times 10^{3} = (0.0180 + x) / ((0.115 - x) \cdot x)\). First, solving the denominator we get: \(0.115x - x^2\). From this, we can figure out that we need to solve equation in a form \(ax^2+bx+c=0\), which is a quadratic equation, where \(a=-1, b=0.115\) and \(c=-1.2 \times 10^{3} \times 0.0180\).
04

- Solve the quadratic equation for x

Solving the quadratic equation \(x^2 - 0.115x + 1.2 \times 10^{3} \times 0.0180 = 0\) with the following formula \(x = {-b \pm \sqrt{b^2-4ac}}/{2a}\), we get two roots. Since the equation is dealing with concentrations, ignore the root that results in a negative concentration of any species.
05

- Substitute x into the equilibrium concentrations

The moles of \(Cl2\) at equilibrium can be obtained by substituting x into the equilibrium concentration of \(Cl2\) which is x. This is because 1 mole of Cl2 is produced for each mole of CO consumed according to the balanced chemical equation.

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Most popular questions from this chapter

For the reaction $$ \mathrm{A}(\mathrm{s}) \rightleftharpoons \mathrm{B}(\mathrm{s})+2 \mathrm{C}(\mathrm{g})+\frac{1}{2} \mathrm{D}(\mathrm{g}) \quad \Delta H^{\circ}=0 $$ (a) Will \(K_{p}\) increase, decrease, or remain constant with temperature? Explain. (b) If a constant-volume mixture at equilibrium at 298 K is heated to 400 K and equilibrium re-established, will the number of moles of \(\mathrm{D}(\mathrm{g})\) increase, decrease, or remain constant? Explain.

Cadmium metal is added to \(0.350 \mathrm{L}\) of an aqueous solution in which \(\left[\mathrm{Cr}^{3+}\right]=1.00 \mathrm{M} .\) What are the concentrations of the different ionic species at equilibrium? What is the minimum mass of cadmium metal required to establish this equilibrium? $$\begin{array}{r} 2 \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{Cd}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Cr}^{2+}(\mathrm{aq})+\mathrm{Cd}^{2+}(\mathrm{aq}) \\ K_{\mathrm{c}}=0.288 \end{array}$$

In one of Fritz Haber's experiments to establish the conditions required for the ammonia synthesis reaction, pure \(\mathrm{NH}_{3}(\mathrm{g})\) was passed over an iron catalyst at \(901^{\circ} \mathrm{C}\) and 30.0 atm. The gas leaving the reactor was bubbled through 20.00 mL of a HCl(aq) solution. In this way, the \(\mathrm{NH}_{3}(\mathrm{g})\) present was removed by reaction with HCl. The remaining gas occupied a volume of 1.82 L at STP. The \(20.00 \mathrm{mL}\) of \(\mathrm{HCl}(\mathrm{aq})\) through which the gas had been bubbled required \(15.42 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Another \(20.00 \mathrm{mL}\) sample of the same HCl(aq) through which no gas had been bubbled required \(18.72 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Use these data to obtain a value of \(K_{\mathrm{p}}\) at \(901^{\circ} \mathrm{C}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})\)

Starting with \(\mathrm{SO}_{3}(\mathrm{g})\) at \(1.00 \mathrm{atm},\) what will be the total pressure when equilibrium is reached in the following reaction at \(700 \mathrm{K} ?\) \(2 \mathrm{SO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.6 \times 10^{-5}\)

In the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\) \(\mathrm{H}_{2}(\mathrm{g}), K=31.4\) at \(588 \mathrm{K} .\) Equal masses of each reactant and product are brought together in a reaction vessel at \(588 \mathrm{K}\). (a) Can this mixture be at equilibrium? (b) If not, in which direction will a net change occur?
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