Question

In a certain chemical reaction, 20.40kJ of heat are evolved. The volume of the reaction is 36.8dm^3 and that of the product is 9.6dm^3. Calculate the change in internal energy for the reaction. Assume the reaction is at ATM pressure.?

Answer 1

But what is atm pressure? It could be anything. I assume you meant normal or standard atm pressure.
dE = q + w
q = -20.40 kJ. Negative because heat is released
w = p*dV.
p = 1 atm
dV = 36.8 - 9.6 = 27.2
Volume decreases meaning that work is done on the system so w is positive. Also the 27.2 is dm^3 * atm. Convert to kJ. Multiply by 101.325
dE = -20.40 + (101.325 x 27.2) = ? kJ.
Check this carefully.