For the conversion of 1 mole of SO2g into SO3g the enthalpy of reaction at constant volume- - U- at 298K is -97-027kJ- Calculate the enthalpy of reaction- - H- at constant pressure-

We have SO_2 (g) #160;+ (1/2)O_2 (g) → SO_3(g) Δ H = Δ U Δ nRT Δ n = 1 1.5= 0.5 Δ H = 97027 0.5*8.314*298 = 98.265 KJ ...

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Standard XII

Physics

Equipartition Theorem

For the conve...

Question

For the conversion of $1$ mole of ${S O}_{2} (g)$ into ${S O}_{3} (g)$ the enthalpy of reaction at constant volume, $Δ U$ , at $298 K$ is $- 97.027 k J$ . Calculate the enthalpy of reaction, $Δ H$ , at constant pressure.

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Solution

We have
$S O_{2} (g) + (1 / 2) O_{2} (g) \to S O_{3} (g)$
$Δ H = Δ U - Δ n R T$
$Δ n = 1 - 1.5 = - 0.5$
$Δ H = - 97027 - 0.5 * 8.314 * 298 = - 98.265 K J$

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For the conversion of 1 mole of $S O_{2} (g)$ into $S O_{3} (g)$ the enthalpy of reaction at constant volume, $Δ U$ at $298 K$ is $- 97.027 K J$ . What is the enthalpy of reaction, $Δ H$ at constant pressure?

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For the conversion of 1 mole of SO2(g) into SO3(g) the enthalpy of reaction at constant volume, ΔU, at 298K is −97.027kJ. Calculate the enthalpy of reaction, ΔH, at constant pressure.

We haveSO2(g)+(1/2)O2(g)→SO3(g)ΔH=ΔU−ΔnRTΔn=1−1.5=−0.5ΔH=−97027−0.5∗8.314∗298=−98.265KJ

For the conversion of $1$ mole of ${S O}_{2} (g)$ into ${S O}_{3} (g)$ the enthalpy of reaction at constant volume, $Δ U$ , at $298 K$ is $- 97.027 k J$ . Calculate the enthalpy of reaction, $Δ H$ , at constant pressure.

We have
$S O_{2} (g) + (1 / 2) O_{2} (g) \to S O_{3} (g)$
$Δ H = Δ U - Δ n R T$
$Δ n = 1 - 1.5 = - 0.5$
$Δ H = - 97027 - 0.5 * 8.314 * 298 = - 98.265 K J$