Consider the reaction at 300K C6H6(l)+152O2(g)→6CO2(g)+3H2O(l):ΔH=−3271KJ what is ∆U for the combustion of 1.5 mole of benzene at 27°C:

ΔH=ΔU+ΔnRT Δn=ngP−ngR =6−152=−32 −3271=ΔU+−32×8.314×300×10−3−3271+3.741=ΔUΔU=−3267.259KJ/ mole For 1.5 mole =1.5×(−3267.259)=4900.88

Consider the reaction at 300K C6H6(l)+152O2(g)→6CO2(g)+3H2O(l):ΔH=−3271KJ what is ∆U for the combustion of 1.5 mole of benzene at 27°C:

ΔH=ΔU+ΔnRT Δn=ngP−ngR =6−152=−32 −3271=ΔU+−32×8.314×300×10−3−3271+3.741=ΔUΔU=−3267.259KJ/ mole For 1.5 mole =1.5×(−3267.259)=4900.88

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Consider the reaction at 300K
$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 {CO}_{2} (g) + 3 H_{2} O (l) : ΔH = - 3271 KJ$ what is $∆$ U for the combustion of 1.5 mole of benzene at 27°C:

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-3274.75 KJ

-4900.88 KJ

-3267.25 KJ

-4906.5 KJ

answer is B.

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Detailed Solution

$ΔH = ΔU + ΔnRT Δn = n_{g} (P) - n_{g} (R) = 6 - \frac{15}{2} = - \frac{3}{2} \begin{array}{l} - 3271 = ΔU + (- \frac{3}{2} \times 8 .314 \times 300 \times 10^{- 3}) \\ - 3271 + 3 .741 = ΔU \\ ΔU = - 3267 .259 KJ / mole \\ For 1 .5 mole = 1 .5 \times (- 3267 .259) = 4900 .88 \end{array}$