For the combustion of 1 mole of liquid benzene at 25oC- the heat of reaction at constant pressure is given by C6H6l - 7 12O2g - 6CO2g - 3H2Ol- U - -780980 calWhat would be the heat of reaction at constant volume-

H=Heat of reaction We have, H = U + n_gRT nbsp; Here, n_g = 6 7.5 = 1.5 Thus, H = U + n_gRT H = 780980 + ( 1.5) × 2 × 298 H = 780090 cal = 780.090 kc ...

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Standard XII

Chemistry

Enthalpy

For the combu...

Question

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

- 780 k c a l

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780 k c a l

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- 580 k c a l

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580 k c a l

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Solution

The correct option is A $- 780 k c a l$
$△ H$ =Heat of reaction
We have, $△ H = △ U + △ n_{g} R T$
Here, $△ n_{g} = 6 - 7.5 = - 1.5$
Thus, $△ H = △ U + △ n_{g} R T$
$△ H = 780980 + (- 1.5) \times 2 \times 298$
$△ H = - 780090 c a l = - 780.090 k c a l$

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Similar questions

Q. 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?

Q. For the combustion of 1 mol of liquid benzene at

25^{o} C

, the heat of reaction at constant pressure is given by,

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

.
What would be the heat of reaction at constant volume?

Q. For combustion of 1 mole of benzene at

25^{0} C

, the heat of reaction at constant pressure is - 780.9 kcal. What will be the heat of reaction at constant volume?

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

Q. For the combustion of 1 mole of a liquid benzene at

25^{o} C

, the enthalpy is given by:

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

cal
What would be the internal energy of the reaction?

Q. For the combustion of 1 mol of liquid benzene at

25^{o} C

, the heat of reaction at constant pressure is given by,

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

.
What would be the heat of reaction at constant volume?

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Enthalpy

Standard XII Chemistry

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For the combustion of 1 mole of liquid benzene at 25oC, the heat of reaction at constant pressure is given by C6H6(l)+712O2(g)→6CO2(g)+3H2O(l); △U=−780980 cal What would be the heat of reaction at constant volume?

The correct option is A −780 kcal△H=Heat of reaction We have, △H=△U+△ngRT Here, △ng=6−7.5=−1.5 Thus, △H=△U+△ngRT △H=780980+(−1.5)×2×298 △H=−780090 cal=−780.090 kcal

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

The correct option is A $- 780 k c a l$
$△ H$ =Heat of reaction
We have, $△ H = △ U + △ n_{g} R T$
Here, $△ n_{g} = 6 - 7.5 = - 1.5$
Thus, $△ H = △ U + △ n_{g} R T$
$△ H = 780980 + (- 1.5) \times 2 \times 298$
$△ H = - 780090 c a l = - 780.090 k c a l$