Answer
(a) $x_{C_2H_2}=0.88$
(b) Low pressures
(c) The number of moles of $C_2H_2$ increases.
Work Step by Step
(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.
