Q1: For the gaseous phase reaction, N₂ + O₂2NO, ΔH = + 80 kJ/kg. mole; the decomposition of NO is favoured by

ANS:B - decrease in temperature.

The reaction you provided is 𝑁2+𝑂2→2𝑁𝑂N2​+O2​→2NO. Given that the enthalpy change (Δ𝐻ΔH) for this reaction is +80 kJ/mol+80kJ/mol, we can see that it's an endothermic reaction because the enthalpy of the products is higher than the enthalpy of the reactants. Now, let's analyze how the factors mentioned affect the decomposition of NO:

1. Increasing the concentration of N2: This would favor the forward reaction, as it would increase the likelihood of collisions between N2 and O2, leading to the formation of more NO. Therefore, this does not favor the decomposition of NO.
2. Decrease in temperature: According to Le Chatelier's principle, decreasing the temperature favors the exothermic direction of the reaction. In an endothermic reaction like the one given, decreasing the temperature would favor the forward reaction, thus forming more NO. So, this does not favor the decomposition of NO.
3. Increase in pressure: Since there are more moles of gas on the reactant side (1 mol of N2 and 1 mol of O2) compared to the product side (2 mol of NO), according to Le Chatelier's principle, an increase in pressure would favor the side with fewer moles of gas. So, increasing the pressure would favor the formation of more NO, thus not favoring the decomposition of NO.
4. Decrease in pressure: Conversely, decreasing the pressure would favor the side with more moles of gas. In this case, it would favor the decomposition of NO, leading to the formation of more N2 and O2.