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schepotkina [342]
3 years ago
5

Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s

tandard conditions, and (NO2) = + 33.2 kJ/mol. Keep one decimal point. 2 NO(g) + O2(g) → 2 NO2(g) ∆Hrxn = –114.2 kJ
Chemistry
1 answer:
Zarrin [17]3 years ago
8 0

Answer:

90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g)  ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

ΔH°f(NO(g)) = 90.3 kJ/mol

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Answer:

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Explanation:

We need to use the formula for heat of vaporization.

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Substitute the values into the formula.

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m is being multiplied by 2260 J/g. The inverse operation of multiplication is division. Divide both sides by 2260 J/g.

\frac{50,000 \ J}{2260 \ J/g} =\frac{2260 \ J/g*m}{2260 \ J/g}

\frac{50,000 \ J}{2260 \ J/g} =m

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\frac{50,000 \ }{2260 \ g} =m

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Round to the nearest whole number. The 1 in the tenth place tells us to leave the number as is.

22 \ g \approx m

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