The difference between - H and - E for the reaction 2C6H6 I - 15O2 g - 12CO2 g - 6H2O I at 250C in KJ is-

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Enthalpy of Combustion

The differenc...

Question

The difference between $Δ H$ and $Δ E$ for the reaction $2 C_{6} H_{6} (I) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (I)$ at $25^{0} C$ in KJ is:

-7.43 KJ

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+3,72 KJ

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-3.72 KJ

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+7.43 KJ

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Solution

The correct option is C -7.43 KJ
Solution:- (A) $- 7.43 k J$
Given reaction-
$2 {C_{6} H_{6}}_{(l)} + 15 {O_{2}}_{(g)} ⟶ 12 {C O_{2}}_{(g)} + 6 {H_{2} O}_{(l)}$
$Δ n_{g} = n_{P} - n_{R} = 12 - 15 = - 3$
Given:- $T = 25 ℃ = (25 + 273) = 298 K$
As we know that,
$Δ H = Δ E + Δ n_{g} R T$
$Δ H - Δ E = Δ n_{g} R T$
$\Rightarrow Δ H - Δ E = - 3 \times 8.314 \times 10^{- 3} \times 300 = - 7.43 k J$
Hence the difference between $Δ H$ and $Δ E$ for the given reaction is $- 7.43 k J$ .

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The difference between ΔH and ΔE for the reaction 2C6H6(I)+15O2(g)→12CO2(g)+6H2O(I) at 250C in KJ is:

The difference between $Δ H$ and $Δ E$ for the reaction $2 C_{6} H_{6} (I) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (I)$ at $25^{0} C$ in KJ is: