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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 a typical equilibrium problem, the value of \(K\) and the initial reaction conditions are given for a specific reaction, and you are asked to calculate the equilibrium concentrations. Many of these calculations involve solving a quadratic or cubic equation. What can you do to avoid solving a quadratic or cubic equation and still come up with reasonable equilibrium concentrations?

A sample of iron(II) sulfate was heated in an evacuated container to \(920 \mathrm{~K}\), where the following reactions occurred: $$ \begin{aligned} 2 \mathrm{FeSO}_{4}(s) & \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(g)+\mathrm{SO}_{2}(g) \\ \mathrm{SO}_{3}(g) & \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \end{aligned} $$ After equilibrium was reached, the total pressure was \(0.836\) atm and the partial pressure of oxygen was \(0.0275\) atm. Calculate \(K_{\mathrm{p}}\) for each of these reactions.

Suppose a reaction has the equilibrium constant \(K=1.7 \times 10^{-8}\) at a particular temperature. Will there be a large or small amount of unreacted starting material present when this reaction reaches equilibrium? Is this reaction likely to be a good source of products at this temperature?

The formation of glucose from water and carbon dioxide is one of the more important chemical reactions in the world. Plants perform this reaction through the process of photosynthesis, creating the base of the food chain: $$ 6 \mathrm{H}_{2} \mathrm{O}(g)+6 \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) $$ At a particular temperature, the following equilibrium concentrations were found: \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=7.91 \times 10^{-2} M,\left[\mathrm{CO}_{2}(g)\right]=\) \(9.3 \times 10^{-1} M\), and \(\left[\mathrm{O}_{2}(g)\right]=2.4 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

Given the following equilibrium constants at \(427^{\circ} \mathrm{C}\), $$ \begin{array}{ll} \mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{1}=2 \times 10^{-25} \\ \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{2}=2 \times 10^{-5} \\ \mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{3}=5 \times 10^{-29} \\ \mathrm{NaO}_{2}(s) \rightleftharpoons \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{4}=3 \times 10^{-14} \end{array} $$ determine the values for the equilibrium constants for the following reactions. a. \(\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) b. \(\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\) c. \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)

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