How much heat is absorbed during production of 112g of NO by the combination of nitrogen and oxygen? . N2(g)+O2(g)→2NO(g), ΔH = + 43 kcal/mol

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Verified Answer :

Alleei Alleei · Answer 1 · 2019-11-29T00:50:45-05:00

Answer : The heat absorbed during the reaction is 160.4 kcal

Explanation :

First we have to calculate the number of moles of NO.

\(\text{Moles of }NO=\frac{\text{Mass of }NO}{\text{Molar mass of }NO}\)

Molar mass of NO = 30 g/mole

\(\text{Moles of }NO=\frac{112g}{30g/mole}=3.73mole\)

Now we have to calculate the heat absorbed during the reaction.

\(\Delta H=\frac{q}{n}\)

or,

\(q=\Delta H\times n\)

where,

\(\Delta H\) = enthalpy change = +43 kcal/mol

q = heat absorb = ?

n = number of moles of NO = 3.73 mol

Now put all the given values in the above formula, we get:

\(q=(+43 kcal/mol)\times (3.73mol)=+160.4kcal\)

Therefore, the heat absorbed during the reaction is 160.4 kcal

JackelineCasarez JackelineCasarez · Answer 2 · 2022-04-01T14:02:38-04:00

The heat absorbed throughout the production of 112g of NO by oxygen and nitrogen in combination:

80.195 kcal

Given equation,

\(N2(g)+O2(g)\) → \(2NO(g),\) Δ\(H = + 43 kcal/mol\)

This shows that 2 moles of NO are required for releasing 43 Kcal. So, total moles of NO can be determined as:

Moles of NO = Mass/Molarity

\(= 112 g/(30 g/mol)\)

\(= 3.73 mol\)

Therefore, the total amount of absorbed heat can be determined as:

⇒ \((43 kcal / 2 mol)\) × \(3.73 mol\)

∵ Heat absorbed = 80.195 kcal

Thus, 80.195 kcal is the correct answer.

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