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

Question

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

I got w= -2.27 x10^3 J.
Is this correct?
If so, I will proceed to the next question and have you check it...
I think if I calculate w wrong, the other answers will be wrong as well so..

Dr.Bob:
I don't understand how you obtained this value for work.
P is 1 atm (STP) and delta V is 1L. Right?

So is the answer -1 atm x 1 L x (1 mol/101.325 atm • L)?

Answer 1

Yes and no. PdV is 1 atm1L = 1 Latm and that is 101.325 J. Now since the volume is changing from 2 L to 1 L that is work done on the system (you had 2L and now just 1 L so the environment had to do work on the system to make that gas decrease in volume) and that means the sign is +. Or if you want to look at it another way w = -pdV = -(1)(-1) = 1 L*atm = +101.325 J.

Remember dE = q+w