For the combustion of 1 mole of liquid benzene at 25-C- the heat of the reaction at constant pressure is given by-C6H6l - 7 12O2 g - 6CO2 g - 3H2O l- H - -780980 calWhat would be the heat of the reaction at constant volume-

We have, Δ H = Δ U + Δ n_g RT Here, Δ n_g = 6 7.5 = 1.5 Thus, Δ U = Δ H – Δ n_g RT Δ U = 780980 ( 1.5) × 2 × 298 Δ U = 780086 cal ...

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Enthalpy

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) + 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 = - 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 = - 780980 k c a l

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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