Answer:
2NO(g) + O2(g) --> 2NO2(g)
now 400 ml of NO × 2 mol of NO2/2 mol of NO
= 400 ml of NO2
now 500 ml of O2 × 2 mol of NO2/1 mol of O2
= 1000 ml of NO2
now 400 ml of NO2 × 1 mol of O2/2 mol of NO
= 200 ml
subtract that from 500 ml of total i.e. 500-200 =300 ml
The total volume of the reaction mixture is 1000 ml -300ml = 700 ml
The enthalpy change : -196.2 kJ/mol
<h3>Further explanation </h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction
2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)
∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

Answer:
Is their a picture with the question?
Explanation: