Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature interval at their values at 298.15 K. Molar heat capacities of NO(g), N2(g), and O2(g) at 298.15 K are 29.86, 29.13, and 29.38 J⋅K−1⋅mol−1. The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

91383 J

Explanation:

The equation of the reaction can be represented as:

    ------>

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

=

where:

= enthalpy of reaction

= the difference in the heat capacities of the products and the reactants.

∴

=

=

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

≅ 91383 J

more info ? ? i don't understand what the question is

explanation:

the answer is 0.8m.

by: nicholas