Reacting 50 mL of H2(g) with 50 mL of C2H4(g) produces 50 mL of C2H6(g) at 1.5 atm. If the reaction produces 3.1 x 10^2 J of heat and the decrease in volume requires the surroundings do 7.6 J of work on the gases, what is the change in internal energy of the gases?

Change in Internal Energy = Q(heat energy) + W(work)

Work = - P x (change in volume)

I know how to apply these equations but have no idea about the " 7.6 J of work" part.

In what sense do you know how to apply the equations if you can't plug in the numbers for them?

ΔU=q+w

The system released energy, so q is negative, thus q=-310 J. Work is done

*on* the system, so w=+7.6 J.

ΔU=(-310)+(+7.6)

ΔU=-302.4 J

Roughly = -300 J (2 sf)

You didn't even need the second equation as far as I understand.

We also know that w=p*(V

_{1}-V

_{2}) (in this case, V

_{2} is smaller than V

_{1} since the volume has decreased, leaving the value inside the brackets as positive and correlating with our positive concept of w in this question, since p is always positive). At constant pressure (1.5 atm=101325*1.5=151,987.5 Pa) we can calculate ΔV in m

^{3}:

V1-V2=w/p

ΔV=V2-V1=-(w/p)

ΔV=-(+7.6/151,987.5)

ΔV=-5*10

^{-5} m

^{3}ΔV=-50 cm

^{3}