# Question faee7

Apr 9, 2015

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

$q = \textcolor{g r e e n}{- \text{2428 J}}$

Likewise, work done by the system means that work was performed by the gas on the surroundings, which once again implies a minus sign.

$w = \textcolor{g r e e n}{- \text{6 kJ}}$

Because the gas burns at constant pressure, the heat given off will also be the enthalpy change, $\Delta H$

$\Delta H = q$ $\to$ 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

$\Delta H = \textcolor{g r e e n}{- \text{2.428 kJ/mol}}$

Now use this equation to determine $\Delta E$

$\Delta E = q + w$

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.