The difference between the heats of reaction at constant pressure and constant volume for the reaction $2 C_{6} H_{6} (l) + 15 O_{2} ⟶ 12 C O_{2} (g) + 6 H_{2} O (l)$ at $25^{\circ} C$ in $k J$ is:

- 7.43

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+ 3.72

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- 3.72

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 7.43

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Solution

The correct option is A $- 7.43$
$2 C_{6} H_{6} (l) + 15 O_{2} ⟶ 12 C O_{2} (g) + 6 H_{2} O (l)$
$Δ H = Δ E + Δ n_{g} R T$
$Δ H - Δ E = Δ n_{g} R T$
$Δ n_{g} = 12 - 15 = - 3$
$Δ H = Δ E = - 3 \times 8.314 \times 298 = - 7432.716 J$

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The difference between the heats of reaction at constant pressure and constant volume for the reaction 2C6H6(l)+15O2⟶12CO2(g)+6H2O(l) at 25∘C in kJ is:

The correct option is A −7.432C6H6(l)+15O2⟶12CO2(g)+6H2O(l)ΔH=ΔE+ΔngRTΔH−ΔE=ΔngRTΔng=12−15=−3ΔH=ΔE=−3×8.314×298=−7432.716J

The difference between the heats of reaction at constant pressure and constant volume for the reaction $2 C_{6} H_{6} (l) + 15 O_{2} ⟶ 12 C O_{2} (g) + 6 H_{2} O (l)$ at $25^{\circ} C$ in $k J$ is:

The correct option is A $- 7.43$
$2 C_{6} H_{6} (l) + 15 O_{2} ⟶ 12 C O_{2} (g) + 6 H_{2} O (l)$
$Δ H = Δ E + Δ n_{g} R T$
$Δ H - Δ E = Δ n_{g} R T$
$Δ n_{g} = 12 - 15 = - 3$
$Δ H = Δ E = - 3 \times 8.314 \times 298 = - 7432.716 J$