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Chapter 13: Problem 100

At a particular temperature, \(K=1.00 \times 10^{2}\) for the reaction $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \leftrightharpoons 2 \mathrm{HI}(g)$$ In an experiment, 1.00 mole of \(\mathrm{H}_{2}, 1.00\) mole of \(\mathrm{I}_{2},\) and 1.00 mole of \(\mathrm{HI}\) are introduced into a \(1.00-\mathrm{L}\) container. Calculate the concentrations of all species when equilibrium is reached.

Short Answer

Expert verified
The equilibrium concentrations of the species are approximately 0.755 M for H₂, 0.755 M for I₂, and 1.49 M for HI.

Step by step solution

01

Write equilibrium expression

To solve the problem, write the equilibrium expression using the definition of K given that: $$K = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}][\mathrm{I}_{2}]}$$
02

Set up Initial, Change, Equilibrium (ICE) table

In this problem, we are given initial concentrations of all species. Write an ICE table to keep track of the initial concentrations, change in concentrations, and equilibrium concentrations for each species. $$ \begin{array}{c|ccc} & [\mathrm{H}_{2}] & [\mathrm{I}_{2}] & [\mathrm{HI}] \\ \hline \text { Initial } & 1.00 & 1.00 & 1.00 \\ \text { Change } & -x & -x & +2 x \\ \text { Equilibrium } & 1.00-x & 1.00-x & 1.00+2 x \\ \end{array} $$
03

Substitute equilibrium concentrations into the equilibrium expression

Substitute the expressions for equilibrium concentrations in terms of x into the equilibrium expression from Step 1: $$K = \frac{(1.00+2x)^2}{(1.00-x)(1.00-x)}$$
04

Substitute the given value of K and solve for x

Substitute the given value of K, which is 1.00 × 10², into the equation from Step 3 and solve for x: $$1.00 × 10^{2} = \frac{(1.00+2x)^2}{(1.00-x)(1.00-x)}$$ You can solve the resulting quadratic equation either by factoring, using the quadratic formula, or using a graphical approach. Solving for x, we get x ≈ 0.245.
05

Calculate the equilibrium concentrations

Substitute x back into the expressions for equilibrium concentrations from the ICE table: $$[\mathrm{H}_{2}]_{eq} = 1.00 - x ≈ 1.00 - 0.245 = 0.755 \ \mathrm{M}$$ $$[\mathrm{I}_{2}]_{eq} = 1.00 - x ≈ 1.00 - 0.245 = 0.755 \ \mathrm{M}$$ $$[\mathrm{HI}]_{eq} = 1.00 + 2x ≈ 1.00 + 2(0.245) = 1.49 \ \mathrm{M}$$ Therefore, the equilibrium concentrations of the species are approximately 0.755 M for H₂, 0.755 M for I₂, and 1.49 M for HI.

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Most popular questions from this chapter

Consider the following reaction: $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$ Amounts of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}, \mathrm{H}_{2},\) and \(\mathrm{CO}_{2}\) are put into a flask so that the composition corresponds to an equilibrium position. If the CO placed in the flask is labeled with radioactive \(^{14} \mathrm{C}\) will \(^{14} \mathrm{C}\) be found only in \(\mathrm{CO}\) molecules for an indefinite period of time? Explain.

An equilibrium mixture contains 0.60 g solid carbon and the gases carbon dioxide and carbon monoxide at partial pressures of 2.60 atm and 2.89 atm, respectively. Calculate the value of \(K_{\mathrm{p}}\) for the reaction $\mathrm{C}(s)+\mathrm{CO}_{2}(g) \Longrightarrow 2 \mathrm{CO}(g)$

At \(25^{\circ} \mathrm{C},\) gaseous \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}(g)\) and \(\mathrm{Cl}_{2}(g)\) to the extent that 12.5\(\%\) of the original \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (by moles) has decomposed to reach equilibrium. The total pressure (at equilibrium) is 0.900 atm. Calculate the value of \(K_{\mathrm{p}}\) for this system.

A 4.72 -g sample of methanol (CH_ 3 \(\mathrm{OH}\) ) was placed in an otherwise empty \(1.00-\mathrm{L}\) flask and heated to \(250 .^{\circ} \mathrm{C}\) to vaporize the methanol. Over time, the methanol vapor decomposed by the following reaction: $$\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)$$ After the system has reached equilibrium, a tiny hole is drilled in the side of the flask allowing gaseous compounds to effuse out of the flask. Measurements of the effusing gas show that it contains 33.0 times as much \(\mathrm{H}_{2}(g)\) as \(\mathrm{CH}_{3} \mathrm{OH}(g) .\) Calculate \(K\) for this reaction at \(250 . \mathrm{C} .\)

The equilibrium constant \(K_{\mathrm{p}}\) is \(2.4 \times 10^{3}\) at a certain temperature for the reaction $$2 \mathrm{NO}(g) \leftrightharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ For which of the following sets of conditions is the system at equilibrium? For those not at equilibrium, in which direction will the system shift? a. $P_{\mathrm{NO}}=0.012 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.11 \mathrm{atm}, P_{\mathrm{O}_{2}}=2.0 \mathrm{atm}$ b. $P_{\mathrm{NO}}=0.0078 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.36 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.67 \mathrm{atm}$ c. $P_{\mathrm{NO}}=0.0062 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.51 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.18 \mathrm{atm}$
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