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Answer 1 · 2015-04-09T01:40:04

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.