## General Chemistry: Principles and Modern Applications (10th Edition)

(a) $x_{C_2H_2}=0.88$ (b) Low pressures (c) The number of moles of $C_2H_2$ increases.
(a) 1. Calculate all the concentrations: $$[CH_4] = ( 0.10 )/(1.00) = 0.10 M$$ - The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_2 ][ H_2 ] ^{ 3 }}{[ CH_4 ] ^{ 2 }}$$ 2. Solve for the missing concentration: $$K_C \times \frac{[ CH_4 ] ^{ 2 }}{[ H_2 ] ^{ 3 }} = [C_2H_2]$$ 3. Evaluate the expression: $$[C_2H_2] = 0.154 \times \frac{( 0.10 )^{ 2 }}{( 0.10 )^{ 3 }}$$ $$[C_2H_2] = 1.5 \space M$$ $$1.00 \space L \times \frac{1.5 \space mol \space C_2H_2}{1 \space L} = 1.5 \space mol \space C_2H_2$$ 4. Calculate the mole fraction of $C_2H_2$: $$x_{C_2H_2}= \frac{1.5 \space mol}{(1.5 + 0.10 + 0.10) \space mol} = 0.88$$ (b) We have 2 moles of gas in the reactants and 4 moles of gas in the products. Therefore, increasing the volume favors the forward reaction (reactants to products), since the products have more volume. Low pressures, at constant temperature, have a larger volume. (c) Increasing the volume favors the forwatd reaction, therefore, $C_2H_2$ will be produced in larger quantities.