Answer : The heat absorbed during the reaction is 160.4 kcal
Explanation :
First we have to calculate the number of moles of NO.
[tex]\text{Moles of }NO=\frac{\text{Mass of }NO}{\text{Molar mass of }NO}[/tex]
Molar mass of NO = 30 g/mole
[tex]\text{Moles of }NO=\frac{112g}{30g/mole}=3.73mole[/tex]
Now we have to calculate the heat absorbed during the reaction.
[tex]\Delta H=\frac{q}{n}[/tex]
or,
[tex]q=\Delta H\times n[/tex]
where,
[tex]\Delta H[/tex] = 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:
[tex]q=(+43 kcal/mol)\times (3.73mol)=+160.4kcal[/tex]
Therefore, the heat absorbed during the reaction is 160.4 kcal
The heat absorbed throughout the production of 112g of NO by oxygen and nitrogen in combination:
80.195 kcal
Given equation,
[tex]N2(g)+O2(g)[/tex] → [tex]2NO(g),[/tex] Δ[tex]H = + 43 kcal/mol[/tex]
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
[tex]= 112 g/(30 g/mol)[/tex]
[tex]= 3.73 mol[/tex]
Therefore, the total amount of absorbed heat can be determined as:
⇒ [tex](43 kcal / 2 mol)[/tex] × [tex]3.73 mol[/tex]
∵ Heat absorbed = 80.195 kcal
Thus, 80.195 kcal is the correct answer.
Learn more about "Nitrogen" here:
brainly.com/question/9176500
