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

-7.43

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3.72

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-3.72

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7.43

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Solution

The correct option is A -7.43
$Δ H = Δ U + Δ n_{g} R T$
$Δ n_{g} = (n_{p r o d u c t})_{g} - (n_{r e a c t a n t})_{g}$
$= 12 - 15 = - 3$
$Δ H - Δ U = - 3 R T$
$= - 3 \times 8.31 \times 298 = - 7429.14 J$
$= - 7.43 K J$

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Similar questions

Q. The difference between heats of reaction at constant pressure and 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) a t 25^{\circ} C

at in kJ is

Q. For

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

the difference in change in enthalpy and change in internal energy at

25^{\circ} C

(kJ) 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

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

25^{\circ} C

k J

is:

The difference between heats of Reaction at constant pressure and constant volume for the Reaction $2 C_{6} H_{6} + 15 O_{2} \to 12 C O_{2} + 6 H_{2} O$ at $25^{\circ} C$ . in $k J$ is

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The difference between heat of reaction at constant pressure and constant volume for the reaction 2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l) at 25∘ C in KJ is

The correct option is A -7.43ΔH=ΔU+ΔngRT Δng=(nproduct)g−(nreactant)g =12−15=−3 ΔH−ΔU=−3RT =−3×8.31×298=−7429.14J =−7.43 KJ

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

The correct option is A -7.43
$Δ H = Δ U + Δ n_{g} R T$
$Δ n_{g} = (n_{p r o d u c t})_{g} - (n_{r e a c t a n t})_{g}$
$= 12 - 15 = - 3$
$Δ H - Δ U = - 3 R T$
$= - 3 \times 8.31 \times 298 = - 7429.14 J$
$= - 7.43 K J$