For the combustion of 1 mole of liquid benzene at 25- C- the heat of the reaction at constant pressure is given by- C 6 H 6 l -7 1-2 O 2 g - 6 CO 2 g -3 H 2 O l - H -780980 calWhat would be the heat of the reaction at constant volume-A- 781080 calB- -780900 calC- 780000 calD- -780086 cal

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For the combu...

Question

For the combustion of 1 mole of liquid benzene at $25^{\circ} C$ , the heat of the reaction at constant pressure is given by,
$C_{6} H_{6} (l) + 7 \frac{1}{2} O_{2} (g) ⟶ 6 C O_{2} (g) + 3 H_{2} O (l)$
$Δ H = - 780980 c a l$
What would be the heat of the reaction at constant volume?

781080 c a l

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- 780900 c a l

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780000 c a l

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- 780086 c a l

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25^{o} C

C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)

Δ H = - 780980 k c a l

25^{o} C,

C_{6} H_{6} (l) + 7 \frac{1}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l); △ U = - 780980 c a l

25^{o} C

C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)

Δ H_{p} = - 780980 c a l m o l^{- 1}

25^{0} C

C_{6} H_{6 (I)} + 7 \frac{1}{2} O_{2 (g)} \to 6 C O_{2 (g)} + 3 H_{2} O_{(l)}

25^{\circ} C

C_{6} H_{6 (1)} + 7 \frac{1}{2} O_{2} (g) \to 6 C O_{2 (g)} + 3 H_{2} O_{(1)}

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