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Consider the chemical equations shown here. Upper N Upper O (g) + Upper O Subscript 3 Baseline (g) right arrow Upper N Upper O Subscript 2 Baseline (g) + Upper O Subscript 2 Baseline (g) delta Upper H Subscript 1 Baseline = negative 198.9 kilojoules. StartFraction 3 Over 2 EndFraction Upper O Subscript 2 Baseline (g) right arrow Upper O Subscript 3 Baseline (g) delta Upper H Subscript 2 Baseline = 142.3 kilojoules. Upper O (g) right arrow one half Upper O Subscript 2 Baseline (g) delta Upper H Subscript 3 Baseline = negative 247.5 kilojoules. What is Delta.Hrxn for the reaction shown below? Upper n upper O (g) plus upper O (g) right arrow upper N upper O subscript 2 (g).

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Answer:

-304.1

Explanation

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Based on the data provided, the overall enthalpy change of reaction is -304.1 kJ.

What is the enthalpy change of the reaction?

The enthalpy change of a reaction is the amount of heat absorbed or given when products are formed from reactants.

The equations of the reactions and accompanying enthalpy changes are given below:

NO (g) + O3 (g) ----> NO2 (g) + O2 (g)  △H = -198.9 kJ ---(i)

03 (g) ---> 3/2 O2 (g) △H = -142.3 kJ

3/2 O2 (g) ---> O3 △H = +142.3 kJ ------ (ii)

O (g) ---> 1/2 O (g) △H = -247.5. -----> (iii)

Adding (i) (ii) and (iii) : NO + O ---> NO2

Overall enthalpy of reaction = - 198.9 kJ + 142.3 kJ - 247.5 =

Overall enthalpy of reaction = -304.1

Therefore, the overall enthalpy change of reaction is -304.1 kJ.

Learn more about enthalpy of reaction at: https://brainly.com/question/14291557

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