Question

Reacting 1L of H2(g) with 1L C2H2(g) (both at STP) results in the formation of 1L of C2H4(g) if the reaction is maintained at the same conditions (STP ang goes to completion. If this reaction produced 6.3kJ of heat (absorbed by surroundings), calculate: a) the PV work b) the change in internal energy of the system c) the change in internal energy of the surroundings.

Answer 1

Solution :-

At STP conditions 1 mol gas = 22.4 L

So

H2(g) + C2H2(g) ----- > C2H4(g)

1 L           1L                           1 L

Change in volume = (1 L) – ( 1 L + 1L) = -1 L

So the a) work done is calculated as

            W= -P*deltaV

                = - 1 atm * (-1L)

               = 1 L atm

1L atm = 101.3 J

So work done = 101.3 J * 1 kJ /1000 J = 0.1013 kJ

b) change in internal energy

delta E = q + w

             = -6.3 kJ + 0.1013 kJ

             = -6.198 kJ

c) change in internal energy of the surrounding

q surrouind = - q system

                      = - (-6.198 kJ)

                      = 6.198 kJ