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Chapter 14: Problem 84

A sample of HI (9.30 \(\times 10^{-3} \mathrm{~mol}\) ) was placed in an empty \(2.00 \mathrm{~L}\) container at \(1000 \mathrm{~K}\). After equilibrium was reached, the concentration of \(\mathrm{I}_{2}\) was \(6.29 \times 10^{-4} \mathrm{M} .\) Calculate the value of \(K_{c}\) at 1000 \(\mathrm{K}\) for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\).

Short Answer

Expert verified
The value of \(K_c\) is 29.0.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation is: \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2 \text{HI}(g) \).
02

Determine initial concentrations

The initial concentration of HI is calculated as follows: \[ \text{Concentration of HI} = \frac{9.30 \times 10^{-3} \text{ mol}}{2.00 \text{ L}} = 4.65 \times 10^{-3} \text{ M} \].
03

Identify equilibrium changes

Let the change in the concentration of \(\text{I}_2\) be \(x\). At equilibrium, \(\text{I}_2\) is \(6.29 \times 10^{-4} \text{ M}\). This means the change \(x\) is \(6.29 \times 10^{-4} \text{ M}\).
04

Calculate equilibrium concentration of HI

The equilibrium concentration of HI will be \((4.65 \times 10^{-3} \text{ M} - 2x)\) at equilibrium: \[ 4.65 \times 10^{-3} - 2(6.29 \times 10^{-4}) = 3.39 \times 10^{-3} \text{ M} \].
05

Write the expression for \(K_c\)

The expression for the equilibrium constant \(K_c\) is: \[ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \].
06

Calculate equilibrium concentration of \(\text{H}_2\) and \(\text{I}_2\)

Since the concentrations of \(\text{H}_2\) and \(\text{I}_2\) both decrease by \(x = 6.29 \times 10^{-4} \text{ M}\), \([\text{H}_2] = 6.29 \times 10^{-4} \text{ M}\) and \([\text{I}_2] = 6.29 \times 10^{-4} \text{ M}\) at equilibrium.
07

Calculate \(K_c\) value

Substitute the equilibrium concentrations into the expression for \(K_c\):\[ K_c = \frac{(3.39 \times 10^{-3})^2}{(6.29 \times 10^{-4})(6.29 \times 10^{-4})} = 29.0 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
In a reversible chemical reaction, chemical equilibrium is achieved when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of the reactants and products remain constant. Think of it like a balanced seesaw, where neither side is going up or down.

Achieving equilibrium doesn't mean the reactions stop. Instead, they continue to occur at the same rate without affecting the overall concentrations. This concept is crucial to understand how reactions behave in closed systems.
  • It is dynamic, not static.
  • Concentrations of reactants and products stay the same.
  • The equilibrium condition is specific to the system's conditions like temperature and pressure.
Chemical Reaction
A chemical reaction involves the transformation of reactants into products. It is depicted through a chemical equation that shows the substances involved and their stoichiometric proportions.

The reversible reaction given: \[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2 \text{HI}(g) \] shows that hydrogen and iodine gas react to form hydrogen iodide, and vice versa.
  • Reversible reactions can proceed in both directions.
  • In equilibrium, concentrations of reactants and products are related through the equation.
  • The coefficients in the equation indicate the ratios in which substances react.
Molar Concentration
Molar concentration, or molarity (M), indicates the amount of a substance present in a specific volume of solution. It's a way to express concentration in chemical equations and reactions, crucial for calculating equilibrium constants.
  • Calculated as moles of solute divided by volume of solution in liters.
  • For HI in the exercise: \[ \text{Concentration} = \frac{9.30 \times 10^{-3} \text{ mol}}{2.00 \text{ L}} = 4.65 \times 10^{-3} \text{ M} \]
  • Concentration changes as reactions proceed toward equilibrium.
Understanding molarity helps in setting up and solving equilibrium expressions, as it's directly used in the calculation of equilibrium constants.
Reaction Kinetics
Reaction kinetics deals with the speed or rate at which chemical reactions occur. It examines how different factors like concentration, temperature, and catalysts affect this rate.

In the case of reaching equilibrium, reaction kinetics explains how the rates of the forward and reverse reactions become equal. Understanding these rates is key to predicting how fast equilibrium can be achieved.
  • The initial speed of reaction depends on concentrations.
  • Temperature changes can significantly alter reaction rates.
  • Kinetics involves analyzing data to formulate rate laws.
Although kinetics focuses on rates, equilibrium discusses the final state where these rates are balanced.

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Most popular questions from this chapter

Calculate the value of the equilibrium constant for the reaction, $$ 4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ given the following equilibrium constants at a certain temperature. $$ \begin{array}{ll} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) & K_{c}=3.2 \times 10^{81} \\ \mathrm{~N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) & K_{c}=3.5 \times 10^{8} \end{array} $$

An equilibrium mixture of \(\mathrm{N}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NH}_{3}\) at \(700 \mathrm{~K}\) contains \(0.036 \mathrm{M} \mathrm{N}_{2}\) and \(0.15 \mathrm{M} \mathrm{H}_{2}\). At this temperature, \(K_{c}\) for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is \(0.29 .\) What is the con- centration of \(\mathrm{NH}_{3}\) ?

A chemical engineer is studying reactions to produce \(\mathrm{SO}_{3}\) as a step in the manufacture of sulfuric acid. The value of \(K_{\mathrm{p}}\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) is \(2.5 \times 10^{10}\) at \(500 \mathrm{~K}\). Will a mixture of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) produce much \(\mathrm{SO}_{3}\) when equilibrium is reached?

At \(1000 \mathrm{~K}, K_{\mathrm{p}}=2.1 \times 10^{6}\) and \(\Delta H^{\circ}=-107.7 \mathrm{~kJ}\) for the reac- tion \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\) (a) A \(0.974\) mol quantity of \(\mathrm{Br}_{2}\) is added to a \(1.00 \mathrm{~L}\) reaction vessel that contains \(1.22 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) gas at \(1000 \mathrm{~K}\). What are the partial pressures of \(\mathrm{H}_{2}, \mathrm{Br}_{2}\), and \(\mathrm{HBr}\) at equilibrium? (b) For the equilibrium in part (a), each of the following changes will increase the equilibrium partial pressure of \(\mathrm{HBr}\). Choose the change that will cause the greatest increase in the pressure of \(\mathrm{HBr}\), and explain your choice. (i) Adding \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) (ii) Adding \(0.10 \mathrm{~mol}\) of \(\mathrm{Br}_{2}\) (iii) Decreasing the temperature to \(700 \mathrm{~K}\).

Baking soda (sodium bicarbonate) decomposes when it is heated: $$ \begin{array}{r} 2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \Delta H^{\circ}=+136 \mathrm{~kJ} \end{array} $$ Consider an equilibrium mixture of reactants and products in a closed container. How does the number of moles of \(\mathrm{CO}_{2}\) change when the mixture is disturbed by the following: (a) Adding solid \(\mathrm{NaHCO}_{3}\) (b) Adding water vapor (c) Decreasing the volume of the container (d) Increasing the temperature
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