Question

Question #faee7

Answer 1

The gas that undergoes combustion produces 2428 J of heat, which implies that `q` will be negative.

`q = color(green)(-"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 = color(green)(-"6 kJ")`

Because the gas burns at constant pressure, the heat given off will also be the enthalpy change, `DeltaH`

`DeltaH = q` `->` at constant pressure;

Convert J to kJ to get

`2428cancel("J") * (10^(-3)"kJ")/(1cancel("J")) = "2.428 kJ"`

Since you're dealing with 1 mole, you can write

`DeltaH = color(green)(-"2.428 kJ/mol")`

Now use this equation to determine `DeltaE`

`DeltaE = q + w`

`DeltaE = -"2.428 kJ" + (-"6 kJ") = color(green)(-"8.428 kJ")`

SIDE NOTE I'll leave the rounding to the correct number of sig figs to you.