Exercises
Exercise 1. Given the following,
2 NO (g) + 2 H2 (g) ⇄ N2 (g) + 2 H2O (g) Kc = 6.7 x 102
what is the equilibrium constant for the following reaction?
1/2 N2 (g) + H2O (g) ⇄ NO (g) + H2 (g) K’c = ?
The equation has been reversed and the coefficients have been divided by 2.
\(\displaystyle K’_c\;=\;\sqrt{\frac{1}{6.7 \times 10^2}}\;=\;\mathbf{3.86 \times 10^{-2}}\)
Exercise 2. Using the following equations,
H+ (aq) + F– (aq) ⇄ HF (aq) Kc = 1.5 x 103
2 H+ (aq) + C2O4-2 (aq) ⇄ H2C2O4 (aq) Kc = 2.6 x 105
2 H+ (aq) + C2O4-2 (aq) ⇄ H2C2O4 (aq) Kc = 2.6 x 105
to find the value of the equilibrium constant for the following reaction.
2 F– (aq) + H2C2O4 (aq) ⇄ 2 HF (aq) + C2O42- (aq) Kc = ?
Multiply the coefficients of the first equation by 2. Then reverse the second equation.
\(\displaystyle 2\;H^+\:(aq)\;+\;2\;F^-\;(aq)\;⇄\;2\;HF\;(aq)\)
\(\displaystyle H_2C_2O_4\;(aq)\;⇄\;2\;H^+\;(aq)\;+\;C_2O_4^{2-}\;(aq)\)