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

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

- R T

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R T

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 2 R T

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 3 R T

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $R T$
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,

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

Suggest Corrections

Similar questions

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

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

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)}

Q. The difference between heats of reaction at constant pressure and 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)

25^{o} C

k J

is:

Join BYJU'S Learning Program

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:

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