Question

#6 The following reaction shows combustion reaction of propane gas at 25 Celsius.

C₃H_8(gas)+5O_2(gas)<===> 3CO_2(gas)+4H₂0_(liq); deltaH=-2219.97 kJ/mol

Initially 1L-chamber contained 2.03M of propane (C₃H₈), 3.50 M of oxygen (O₂) and 0.030M of carbon dioxide (CO₂). After equilibrium is reached, the chamber contained 1.53 M of propane (C₃H₈).

a)Using the information given,calculate equlibrium constant (K_c) of thid reaction.

b) After equilibrium is reached, if the chamber size is increased to 5L, predict the direction of equilibrium shift. Explian your answer.

c)After equilibrium is reached, if we increase temperature to 1000 Celsius, predict the direction of equilibrium shift. Explain your answer.

Answer 1

a) C3H8(gas) + 5O2(gas) <------> 3CO2(g) + 4H2O(l)

Kc = [CO2]^3/[C3H8][O2]^5

at equillibrium

[ C3H8 ] = 2.03 -x

[ O2] = 3.50 - 5x

[ CO2] = 0.030 + 3x

Kc = (0.030 + 3x)^3/(2.03 - x) (3.50 - 5x)^5

at equillibrium

[C3H8 ] = 1.53M

2.03 - x = 1.53M

x = 0.5

Therefore,

Kc = (0.030+3x)^3/(2.03 - x) (3.50-5x)^5

= (0.030 + 3(0.5))^3/(2.03 - 0.5 ) ( 3.50 -5(0.5))^5

= 3.58/ 1.53

= 2.34

b) Increasing volume shift the equillibrium to the side which higher mole of gaseous product , reactant side having more moles (6 moles ( 1C3H8 + 5O2) ) than the product side (3mol (3CO2)).So, the equillibrium shift to the reactant side.

c) This reaction is exothermic reaction ( heat is released in this reaction).Exothermic reaction is not favourable for product side. So , the equillibrium shift reactant side.