The difference in the heat of reaction at constant volume and at constant pressure for the given reaction is:

$C H_{4} (g) + O_{2} (g) M n^{2 +} - -- \to H C O O H (l) + H_{2} (g)$

–RT

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-2RT

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– 3RT

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Solution

The correct option is B RT
We know that the heat of reaction at a constant volume is given by the internal energy of the system whereas the heat of reaction at a constant pressure is given by the enthalpy of the system.
Again,

$Δ H = Δ U + Δ n_{g} R T$
$\Rightarrow Δ U - Δ H = - Δ n_{g} R T$
$\Rightarrow Δ U - Δ H = - (1 - 2) R T$
$\Rightarrow Δ U - Δ H = R T$

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Similar questions

The difference in the heat of reaction at constant volume and at constant pressure for the given reaction is:

$C H_{4} (g) + O_{2} (g) M n^{2 +} - -- \to H C O O H (l) + H_{2} (g)$

Q. For the combustion of

C_{3} H_{8} (g)

at constant temperature

Δ H - Δ U

is:

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

The difference between heat of reaction at constant pressure and constant volume for the reaction given below at 25 $^{0}$ C in kJ is: $2 C_{6} H_{6 (l)} + 15 O_{2 (g)} \to 12 C O_{2 (g)} + 6 H_{2} O_{(l)}$

Q. The difference between the heat of reaction at constant pressure and constant volume for the reaction given below at

25^{o} C

in KJ is:

2 C_{6} H_{6 (l)} + 15 O_{2 (g)} ⟶ 12 {C O}_{2 (g)} + 6 H_{2} O_{(l)}

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The difference in the heat of reaction at constant volume and at constant pressure for the given reaction is: CH4(g)+O2(g)Mn2+−−−→HCOOH(l)+H2(g)

The difference in the heat of reaction at constant volume and at constant pressure for the given reaction is:

$C H_{4} (g) + O_{2} (g) M n^{2 +} - -- \to H C O O H (l) + H_{2} (g)$