Answer:
W=-37.6kJ, therefore, work is done on the system.
Explanation:
Hello,
In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

Next, the total moles:

After that, since the process is isobaric, we can compute the work as:

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

Thereby, the magnitude and direction of work turn out:

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).
Regards.
Answer:
The ΔHrxn for the above equation = 179 kJ/mol
Explanation:
The reaction bond enthalpies are for the reactant;
3 × N-H = 3 × 390 = 1,170 kJ/mol
2 × O=O = 2 × 502 = 1004 kJ/mol
The reaction bond enthalpies are for the product;
3 × N-O = 3 × 201 = 603 kJ/mol
3 × O-H = 3 × 464 = 1,392 kJ/mol
The ΔHrxn for the above equation is therefore;
ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol
Li because its charge is +1.