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Problem Solving and K c

In this study guide we will solve problems either to calculate the equilibrium constant or to determine the equilibrium concentrations of reactants and products once the system is at equilibrium. To calculate K c we would need at least one of the equilibrium concentrations. If we are given initial concentrations and the value of K c or K p , we can calculate the equilibrium concentrations.

Calculate K c if Given One or More Equilibrium Concentrations

If we are given equilibrium concentrations, we can calculate K c . For example, consider the following reaction:

H 2 O (g) + CH 4 (g) ⇄ CO (g) + 3 H 2 (g)     K c = ? at 1127 o C

The equilibrium concentrations are: [H 2 O] eq = 0.038 M, [CH 4 ] eq = 0.045 M, [CO] eq = 0.18 M, and [H 2 ] eq = 0.35 M

Write the equilibrium constant expression. \(\displaystyle \frac{[CO][H_2]^3}{[H_2O][CH_4]}\;=\;K_c\)

Now we plug the equilibrium concentrations into K c

\(\displaystyle \frac{(0.18)(0.35)^3}{(0.038)(0.045)}\;=\;4.5\)   The value of K c is 4.5. The products are favored at equilibrium. Notice that H 2 O is included in the equilibrium constant expression because it is a gas.

In this next problem, we are given two initial concentrations and one equilibrium concentration. In this case we will learn how to set up an ICE table . First, consider the following reaction at a certain temperature:

2 Fe 3+ (aq) + 3 I – (aq) ⇄ 2 Fe 2+ (aq) + I 3 – (aq)    K c = ?

What is K c if the initial concentrations of Fe 3+ and I – are 0.250 M and 0.350 M respectively. The equilibrium concentration of I 3 – is 0.0752 M.

First, make sure the chemical equation is balanced, and then write the equilibrium constant expression.

\(\displaystyle K_c\;=\;\frac{[Fe^{2+}]^2[I_3^-]}{[Fe^{3+}]^2[I^-]^3}\)

Now, we set up the ICE table.

The I stands for initial concentrations (or pressures), C is the change in concentration as the reaction proceeds, and E represents the equilibrium concentrations. The initial concentrations of Fe 3+ and I – are 0.250 M and 0.350 M respectively. The initial concentration of Fe 2+ and I 3 – is equal to zero. These concentrations go into the first row of the ICE table. As the reaction proceeds the concentration of the reactants will decrease by some amount x while the concentrations of products will increase by some amount x. Make sure you use the stoichiometric coefficients in the ICE table with x. This goes into the second row of the table. In the third row of the table write the equilibrium concentrations in terms of x.

In the problem we are told the equilibrium concentration of I 3 – is 0.0752 M. From the ICE table, we see 0.0752 M is equal to x. Because we know x, we can determine the equilibrium concentrations of all species. Below is the ICE table with calculations.

Now we have all of the equilibrium concentrations, we can plug them into the K c expression.

\(\displaystyle K_c\;=\;\frac{[Fe^{2+}]^2[I_3^-]}{[Fe^{3+}]^2[I^-]^3}\;=\;\frac{(0.1504)^2(0.0752)}{(0.0996)^2(0.124)^3}\;=\;89.9\)

  The value of K c is 89.9. At equilibrium, products predominate.

Calculation of Equilibrium Concentrations if Given K c and Initial Concentrations

Consider the following equation with K c = 2.5 x 10 2 at some temperature.

H 2 (g) + Br 2 (g) ⇄ 2 HBr (g)

The initial concentrations are: [H 2 ] = 0.465 M and [Br 2 ] = 0.465 M. Calculate the equilibrium concentrations of all species.

First, make sure the chemical equation is balanced. Next, write the K c expression.

\(\displaystyle K_c\;=\;\frac{[HBr]^2}{[H_2][Br_2]}\;=\;2.5 \times 10^2\)   We then set up the ICE table. The initial concentrations of H 2 and Br 2 are 0.465 M. The initial concentration of HBr is equal to zero. These concentrations go into the first row of the ICE table. As the reaction proceeds, the concentrations of H 2 and Br 2 decrease by some amount x. The concentration of HBr increases by some amount 2x. The coefficients in the chemical equation are always used. In the third row, the equilibrium concentrations in terms of x are written.

Next, put the equilibrium concentrations into the equilibrium constant expression.

\(\displaystyle \frac{(2x)^2}{(0.465\;-\;x)(0.465\;-\;x)}\;=\;2.5 \times 10^2\)   The expression needs to be solved for x. First, we can simplify the expression as:

\(\displaystyle \frac{4x^2}{(0.465\;-\;x)^2}\;=\;2.5 \times 10^2\)   The left side is a perfect square, and we can take the square root of both sides of the equation.

\(\displaystyle \sqrt{\frac{4x^2}{(0.465\;-\;x)^2}}\;=\;\sqrt{2.5 \times 10^2}\)   \(\displaystyle \frac{2x}{(0.465\;-\;x)}\;=\;15.8\)

The equation can now be solved for x.

\(\displaystyle 2x\;=\;15.8\;(0.465\;-\;x)\)   \(\displaystyle 2x\;=\;7.347\;-\;15.8x\)   \(\displaystyle 17.8x\;=\;7.347\)   \(\displaystyle x\;=\;\frac{7.347}{17.8}\;=\;0.413\;M\)   Now, we can calculate the equilibrium concentrations of each substance in the reaction using the equilibrium concentrations in the third row of the ICE table.

[H 2 ] eq = [Br 2 ] eq = 0.465 M – x = 0.465 M – 0.413 M = 0.052 M

[HBr] eq = 2x = 2 x 0.413 M = 0.826 M

By realizing we had a perfect square, we did not have to use the quadratic formula to solve for x. When both reactants have the same initial concentrations, we usually will have a perfect square as long as the stoichiometric coefficient of the product is equal to two. In the next problem, we will need to use the quadratic formula.

Using Q c and the Quadratic Formula to Solve for Equilibrium Concentrations

Consider the following reaction at a certain temperature.

I 2 (g) + H 2 (g) ⇄ 2 HI (g)   K c = 55.2

The initial concentrations are: [I 2 ] = 0.0030 M, [H 2 ] = 0.0050 M, [HI] = 0.030 M. Calculate the equilibrium concentrations of all species.

For this problem we are given initial concentrations of both reactants and product. We need to determine which way the reaction will proceed to reach equilibrium. We can calculate Q c and then compare it to K c .

\(\displaystyle Q_c\;=\;\frac{[HI]^2}{[I_2][H_2]}\;=\;\frac{(0.30)^2}{(0.0030)(0.0050}\;=\;60\)   Q c > K c , and the reaction will proceed to the left to reach equilibrium. When we set up the ICE table, remember the reaction will proceed from right to left. The reactants will increase in concentration while the product will decrease in concentration.

Next, write the equilibrium constant expression.

\(\displaystyle K_c\;=\;\frac{[HI]^2}{[I_2][H_2]}\;=\;55.2\)   \(\displaystyle \frac{(0.030\;-\;2x)^2}{(0.0030\;+\;x)(0.0050\;-\;x)}\;=\;55.2\)   Here, we do not have a perfect square so we will have to solve the quadratic formula. First, multiply the terms in the equation and then collect the like terms.

\(\displaystyle \frac{4x^2\;-\;0.12x\;+\;9.0\times10^{-4}}{x^2\;+\;0.0080x\;+\;1.5\times 10^{-5}}\;=\;55.2\)   \(\displaystyle 4x^2\;-\;0.12x\;+\;9.0\times10^{-4}\;=\;55.2(x^2\;+\;0.0080x\;+\;1.5\times 10^{-5})\)   \(\displaystyle 4x^2\;-\;0.12x\;+\;9.0\times10^{-4}\;=\;-55.2x^2\;+\;0.4416x\;+\;8.28\times 10^{-4}\)

Collect all terms on one side of the equation and set the equation equal to zero.

\(\displaystyle 51.2x^2\;+\;0.5616x\;+\;7.2\times 10^{-5}\;=\;0\)

We have an equation in the form of ax 2 + bx + c = 0 where a = 51.2, b = 0.5616, and c = 7.2 x 10 -5 .

The quadratic formula is: \(\displaystyle x\;=\;\frac{-b\pm\sqrt{b^2\;-\;4ac}}{2a}\)

\(\displaystyle x\;=\;\frac{-0.5616\pm\sqrt{0.5616^2\;-\;4(51.2)(-7.2\times 10^{-5})}}{2\times 51.2}\)   \(\displaystyle x\;=\;\frac{-0.5616 \pm 0.5746}{102}\)   There are two values of x. x = 1.27 x 10 -4 M and x = -0.111 M.

The negative value does not work because we cannot have negative concentrations. The value of x is equal to 1.27 x 10 -4 M. Now, we can calculate the equilibrium concentrations.

[I 2 ] eq = 0.0030 + x = 0.0030 + (1.27 x 10 -4 M) = 0.0031 M [H 2 ] eq = 0.0050 + x = 0.0050 + (1.27 x 10 -4 M) = 0.0051 M [HI] eq = 0.030 + 2x = 0.030 + 2(1.27 x 10 -4 M) = 0.030 M

As you can see, the equilibrium constant expressions can get complex depending on the equation stoichiometry. In the next Study Guide, you will be shown how to use a simplifying assumption which will simplify the math.

Watch the Following Videos

Exercise 1. Calculate K p for the following reaction, at 600 o C, if the equilibrium pressures are P PH 3 = 5.4 x 10 -4 atm, P P 2 = 0.426 atm, and P H 3 = 0.795 M

2 PH 3 (g) ⇄ P 2 (g) + 3 H 2 (g)

Check Answer/Solution to Exercise 1

Exercise 2. The following reaction has K c = 32 at 227 o C.

2 BrCl (g) ⇄ Br 2 (g) + Cl 2 (g)

The initial concentrations are [BrCl] = 0.0450 M, [Br 2 ] = 0.0300 M, and [Cl 2 ] = 0.025 M. Calculate the equilibrium concentrations of all species.

Check Answer/Solution to Exercise 2

Exercise 3. Consider the following reaction at 500 o C.

I 2 (g) + H 2 (g) ⇄ 2 HI (g)

K c is equal to 54.9. Calculate all the equilibrium concentrations if the initial concentrations of I 2 and H 2 are 0.045 M and 0.035 M, respectively.

Check Answer/Solution to Exercise 3

Exercise 4. The equilibrium constant, K p for the reaction below is 49.

H 2 (g) + I 2 (g) ⇄ 2 HI (g)

At 495 K, flask was filled with H 2 and I 2 both with a pressure of 3.6 atm. Calculate the pressures of all species at equilibrium.

Check Answer/Solution to Exercise 4

Exercise 5. The following reaction has K c = 4.60 x 10 -3 at a certain temperature.

N 2 O 4 (g) ⇄ 2 NO 2 (g)

The initial concentration of N 2 O 4 was 0.425 M. What are the concentrations of N 2 O 4 and NO 2 at equilibrium?

Check Answer/Solution to Exercise 5

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