Consider the reaction below: CH4(g)+O2(g)→ CO2(g)+2 H2O(g)
ΔH=−890.4kJ

Calculate the amount of heat (q) produced by the combustion of 3.04 L of methane at 1.52 atm under STP conditions.

Include your sign indicating if heat is absorbed or released.

the change in internal energy of the gases is  -3.0924 KJ

Explanation:

The computation of the change in internal energy of the gases is as follows;

Given that

Q = -3.1 KJ

= -3100 J

It is negative as the heat is lost from the mixture

And,

W = + 7.6 J

It is positive as the work is done in the mixture

Now we use the following equation

Delat E = Q + W

= -3100 J + 7.6 J

= -3092.4 J

= -3.0924 KJ

Hence, the change in internal energy of the gases is  -3.0924 KJ