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Chapter 13: Problem 67

In which direction will the position of the equilibrium $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ be shifted for each of the following changes? a. \(\mathrm{H}_{2}(g)\) is added. b. \(\mathrm{I}_{2}(g)\) is removed. c. \(\mathrm{HI}(g)\) is removed. d. In a rigid reaction container, some \(\operatorname{Ar}(g)\) is added. e. The volume of the container is doubled. f. The temperature is decreased (the reaction is exothermic).

Short Answer

Expert verified
a. The equilibrium will shift to the right. b. The equilibrium will shift to the right. c. The equilibrium will shift to the left. d. The equilibrium will not shift. e. The equilibrium will shift to the right. f. The equilibrium will shift to the right.

Step by step solution

01

Determine the effect of the change

Increasing the concentration of a reactant (H2) will cause the system to move towards the product side (to the right). Therefore, the equilibrium will shift to the right to minimize the effect of the added H2. #b. Removing I2(g)#
02

Determine the effect of the change

Decreasing the concentration of one of the products (I2) will cause the system to replenish the depleted product, thus shifting the equilibrium to the product side (to the right). #c. Removing HI(g)#
03

Determine the effect of the change

Decreasing the concentration of a reactant (HI) will cause the system to shift towards the reactants side (to the left) in order to minimize the effect and increase the concentration of HI. #d. Adding Ar(g) in a rigid container#
04

Determine the effect of the change

Adding an inert gas (Ar) that doesn't participate in the reaction, in a rigid container (at constant volume) will not cause a change in the partial pressures of the reactants or products. Therefore, the equilibrium will not shift. #e. Doubling the volume of the container#
05

Determine the effect of the change

Doubling the volume of the container results in a decrease in the partial pressure of all gases in the system. To minimize this effect, the equilibrium will shift in the direction that increases the number of moles of gas. In this case, the equilibrium will shift to the right (as there are 2 moles of HI for every mole of H2 and I2). #f. Decreasing the temperature (reaction is exothermic)#
06

Determine the effect of the change

If the reaction is exothermic, it releases heat as it proceeds towards the products. When the temperature is decreased, the system will try to minimize the effect by producing more heat. Consequently, the equilibrium will shift to the right, favoring the formation of products.

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Most popular questions from this chapter

For the following reaction at a certain temperature $$ \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g) $$ it is found that the equilibrium concentrations in a 5.00-L rigid container are \(\left[\mathrm{H}_{2}\right]=0.0500 M,\left[\mathrm{~F}_{2}\right]=0.0100 M\), and \([\mathrm{HF}]=\) \(0.400 \mathrm{M}\). If \(0.200 \mathrm{~mol} \mathrm{~F}_{2}\) is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

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.

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C}\), equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

Peptide decomposition is one of the key processes of digestion, where a peptide bond is broken into an acid group and an amine group. We can describe this reaction as follows: Peptide \((a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) acid group \((a q)+\) amine group \((a q)\) If we place \(1.0\) mol peptide into \(1.0 \mathrm{~L}\) water, what will be the equilibrium concentrations of all species in this reaction? Assume the \(K\) value for this reaction is \(3.1 \times 10^{-5}\).

The equilibrium constant is \(0.0900\) at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ For which of the following sets of conditions is the system at equilibrium? For those which are not at equilibrium, in which direction will the system shift? a. A \(1.0\) - \(\mathrm{L}\) flask contains \(1.0 \mathrm{~mol} \mathrm{HOCl}, 0.10 \mathrm{~mol} \mathrm{Cl}_{2} \mathrm{O}\), and \(0.10 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) b. A 2.0-L flask contains \(0.084 \mathrm{~mol} \mathrm{HOCl}, 0.080 \mathrm{~mol} \mathrm{Cl}_{2} \mathrm{O}\), and \(0.98 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) c. A 3.0-L flask contains \(0.25 \mathrm{~mol} \mathrm{HOCl}, 0.0010 \mathrm{~mol} \mathrm{Cl}_{2} \mathrm{O}\), and \(0.56 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\)
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