Answer to Question #185561 in General Chemistry for mike

Question

A mixture was prepared using 1.00mol of propanoic acid, 2.00mol of ethanol and 5.00mol of water. At a given temperature, the mixture was left to reach equilibrium according to the following equation.

CH3CH2COOH  +  CH3CH2OH     CH3CH2COOCH2CH3  +  H2O       ∆Hο = –22kJ mol–1

The equilibrium mixture contained 0.54mol of the ester ethyl propanoate.

(a)(i). Calculate the amounts, in moles, of propanoic acid, of ethanol and of water in this equilibrium mixture.

(ii). Write an expression for the equilibrium constant, Kc, for this equilibrium.

(iii). Calculate a value for Kc for this equilibrium at this temperature. Explain why this Kc value has no units.

(b). For this equilibrium, predict the effect of an increase in temperature on each of the following.

(i). the amount, in moles, of ester at equilibrium

(ii). the time taken to reach equilibrium

(iii). the value of Kc

Answer 1

(a)

i) . moles of propanoic acid = 1 mol - 0.54 mol

= 0.46 mol

moles of ethanol = 2 mol - 0.54 mol

= 1.46 mol

moles of water = 5 mol + 0.54 mol

= 5.54 mol

ii) . K_c = $\frac{[CH_3CH_2COOCH_2CH_3][ H_2O]}{[CH_3CH_2COOH][CH_3CH_2OH]}$

iii). K_c = $\frac{[0.54][5.54]}{[0.46][1.46]}$ = 4.45

there are no units as equal to no. of moles on each side of the equation

(b) .

i). Decreased

( forward reaction is exothermic (-22kJ) so will move to endothermic side )

ii). Decreased

( rate increases so time decreases )

iii). Decreased

( equilibrium position shift left so K_c gets smaller )