Exercises
Exercise 1. Calculate Kp for the following reaction, at 600 oC, if the equilibrium pressures are PPH3 = 5.4 x 10-4 atm, PP2 = 0.426 atm, and PH3 = 0.795 M
\(\displaystyle K_p\;=\;\frac {P_P\times P_{H_2}^{3}}{P_{PH_3}^2}\)
\(\displaystyle K_p\;=\;\frac {0.426\times (0.795)^3}{(5.4\times 10^{-4})^2}\;=\;\bf{7.3\times10^5}\)
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Exercise 2. The following reaction has Kc = 32 at 227oC.
2 BrCl (g) ⇄ Br2 (g) + Cl2 (g)
The initial concentrations are [BrCl] = 0.0450 M, [Br2] = 0.0300 M, and [Cl2] = 0.025 M. Calculate the equilibrium concentrations of all species.
Set up ICE table
Only the initial concentrations could be put into the ICE table. We are given the concentrations of both products and reactants. We need to calculate Q to see which way the reaction proceeds to equilibrium.
\(\displaystyle Q_c\;=\;\frac {(0.0300)(0.025)}{(0.0450)^2}\;=\;0.37\)
Qc < Kc The reaction will proceed to the right. Now we can finish the ICE table.
\(\displaystyle Q_c\;=\;\frac{(0.025\;+\;x)(0.0300\;+\;x)}{(0.0450\;-\;2x)^2}\;=\;32\)
Multiply and then collect all like terms.
\(\displaystyle\frac{(0.0300+x)(0.025+x)}{(0.0450-2x)(0.0450-2x)}\;=\;32\)
\(\displaystyle x^2+0.055x+(7.5\times 10^{-4}=32(4x^2-0.18x+0.002025)\)
Collect all like terms on one side of the equation, and then set the equation equal to zero.
\(\displaystyle 127x^2-5.815x+0.06405=0\)
Use the quadratic formula to solve for x.
\(\displaystyle \frac {-b\pm\sqrt{b^2-4ac}}{2a}\)
a = 127, b = -5.815, c = 0.0645
\(\displaystyle \frac {-(-5.815)\pm\sqrt{(-5.815)^2-4\times 127\times0.0645}}{2\times 127}\)
\(\displaystyle \frac{5.815\pm 1.13}{254}\)
x = 0.0273 M and x = 0.0184 M The value of x cannot give a negative concentration which 0.0273 M would, so x = 0.0184.
The equilibrium concentrations are:
[Cl2]eq = 0.025 M + x = 0.025 M + 0.0184 M = 0.043 M
[Br2]eq = 0.0300 M + x = 0.0300 M + 0.0184 M = 0.0484 M
[BrCl]eq = 0.0450 M – 2x = 0.0450 M – 2 x 0.0184 M = 0.0082 M
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Exercise 3. Consider the following reaction at 500 oC.
I2 (g) + H2 (g) ⇄ 2 HI (g)
Kc is equal to 54.9. Calculate all the equilibrium concentrations if the initial concentrations of I2 and H2 are 0.045 M and 0.035 M, respectively.
\(\displaystyle K_c=\frac{[HI]^2}{[I_2][H_2]}=54.9\)
Set up the ICE table.
\(\displaystyle \frac{[2x]^2}{[0.045-x][0.035-x]}=54.9\)
Multiply and collect all like terms.
\(\displaystyle 50.9x^2-4.392x+0.086468=0\)
Use the quadratic formula. a = 50.9, b = -4.392, c = 0.086468
\(\displaystyle \frac {-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\displaystyle \frac {-(-4.392)\pm\sqrt{(-4.392)^2-4(50.9)(0.086468}}{2(50.9)}=\frac{4.392\pm 1.298}{101.8}\)
x = 0.0559 M, x = 0.0304 M
x cannot be 0.0559 M because it is larger than either initial concentration. x = 0.0304 M
[I2]eq = 0.035 M – 0.0304 M = 0.0046 M
[H2]eq = 0.045 M – 0.0304 M = 0.0146 M
[HI]eq = 2x = 2 x 0.0304 M = 0.0608 M
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Exercise 4. The equilibrium constant, Kp for the reaction below is 49.
H2 (g) + I2 (g) ⇄ 2 HI (g)
At 495 K, flask was filled with H2 and I2 both with a pressure of 3.6 atm. Calculate the pressures of all species at equilibrium.
\(\displaystyle K_c=\frac{[HI]^2}{[I_2][H_2]}=49\)
Set up the ICE table.
\(\displaystyle \frac{(2x)^2}{(3.6-x)(3.6-x)}=49\)
\(\displaystyle \frac{(2x)^2}{(3.6-x)^2}=49\)
We have a perfect square. Take the square root of both sides and solve for x.
\(\displaystyle \sqrt{\frac{(2x)^2}{(3.6-x)^2}}=\sqrt{49}\)
\(\displaystyle \frac{2x}{3.6-x}=49\)
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Exercise 5. The following reaction has Kc = 4.60 x 10-3 at a certain temperature.
N2O4 (g) ⇄ 2 NO2 (g)
The initial concentration of N2O4 was 0.425 M. What are the concentrations of N2O4 and NO2 at equilibrium?