When 1mol of a gas burns at constant pressure, it produces 2428 j of heat and does 6 kj of work?

The gas that experiences combustion yields 2428 J of heat, which suggests that q will be negative.

q=−2428 J

Similarly, work done by the system means that work was done by the gas on the surrounds, which once again suggests a minus sign.

w=−6 kJ

Since the gas burns at constant pressure, the heat given off will also be the enthalpy change, ΔH

ΔH=q → at constant pressure;

Change J to kJ to get

2428J x 10^−3kJ/ 1J = 2.428 kJ

Since you're distributing with 1 mole, you can write:

ΔH=−2.428 kJ/mol

Now use this equation to determine ΔE

ΔE=q+w

ΔE=−2.428 kJ+(−6 kJ)=−8.428 kJ

c. magnitude and direction