Question

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

Answer 1

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

Answer 2

The gas that undergoes combustion produces 2428 J of heat, which implies that
q will be negative.
q =â’ 2428 J
Likewise, work done by the system means that work was performed by the gas on the surroundings, which once again implies a minus sign.
w =â’6 kJ
Because the gas burns at constant pressure, the heat given off will also be the enthalpy change, Δ H
ΔH = q → at constant pressure;
Convert J to kJ to get
2428Jâ‹…10-3k/1J =2.428 kJ
Since you're dealing with 1 mole, you can write
Δ H = ⒠2.428 kJ/mol
Now use this equation to determine Δ E
Δ E = q + w
Δ E = â’ 2.428 kJ + (â’6 kJ) = â’ 8.428 kJ
SIDE NOTE I'll leave the rounding to the correct number of sig figs to you