Chemistry (4th Edition)

b) $[ A ] = 0.41 M$ $[ B ] = 0.41 M$ $[ C ] = 0.054 M$
- The new reaction is: $$2C \leftrightharpoons A + B$$ So, the new equilibrium constant is: $$K_c = \frac{1}{1.7 \times 10^{-2}} = 59$$ 1. Write the equilibrium constant expression: - The exponent of each concentration is equal to its balance coefficient. $$K_c = \frac{[Products]}{[Reactants]} = \frac{[ A ][ B ]}{[ C ]^{ 2 }}$$ 2. At equilibrium, these are the concentrations of each compound: $[ C ] = 0.875 \space M - 2x$ $[ A ] = x$ $[ B ] = x$ $$59 = \frac{(x)(x)}{(0.875-2x)^2}$$ x $\approx$ 0.4107 (in order to get a positive concentration of C) 3. $[ C ] = 0.875 \space M - 2(0.4107) M= 0.054 M$ $[ A ] = 0.41 M$ $[ B ] = 0.41 M$