Consider the reaction below:

N2(g) + 2 O2(g) 2 NO2(g) ΔH = +66.4 kJ

What would the enthalpy change for the following reaction be?

4 NO2(g) 2 N2(g) + 4 O2(g) ΔH = ???

a. +66.4 kJ
b. -66.4 kJ
c. +132.8 kJ
d. -132.8 kJ
e. -33.4 kJ

D.

Explanation:

The forward reaction is endothermic. This is shown by the positive enthalpy change (+ 66.4 kJ). The reverse reaction will, therefore, be exothermic (-66.4 kJ).

However, because in the reverse reaction we are dealing with double the moles as in the forward reaction, the enthalpy change will also be double.

66.4 * 2 = 132.8

= - 132.8 kJ

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

no, the red giant is the hottest with 2.00.000.000 degrees celsius. while the white dwarf only 25,000 degrees celsius.