The difference between -H and -E on a molar basis for the combustion of n-octane at 25-C would be-

The correct option is C 11.15 kJ H E= 4.5× 8.315× 298J= 11.5kJ Balanced chemical equation for combustion of n octane 2 C _ 8 H _ 18 ( s ) +25 O _ 2 ...

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Standard XII

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The differenc...

Question

The difference between $Δ$ H and $Δ$ E (on a molar basis) for the combustion of n-octane at $25^{\circ}$ C would be:

-13.6 kJ

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-1.14 kJ

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-11.15 kJ

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+11.15 kJ

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Solution

The correct option is C -11.15 kJ
$△ H - △ E = - 4.5 \times 8.315 \times 298 J = - 11.5 k J$
Balanced chemical equation for combustion of n-octane
$2 C_{8} H_{18} (s) + 25 O_{2} (g) ⟶ 16 C O_{2} (g) + 18 H_{2} O (l) △ n = \frac{16 - 25}{2} = \frac{- 9}{2} = - 4.5 R = 8.315 J {m o l}^{- 1} K^{- 1} (c o n s t a n t) T = 298 K (g i v e n)$

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The difference between ΔH and ΔE (on a molar basis) for the combustion of n-octane at 25∘C would be:

The correct option is C -11.15 kJ△H−△E=−4.5×8.315×298J=−11.5kJBalanced chemical equation for combustion of n-octane2C8H18(s)+25O2(g)⟶16CO2(g)+18H2O(l)△n=16−252=−92=−4.5R=8.315Jmol−1K−1(constant)T=298K(given)

The difference between $Δ$ H and $Δ$ E (on a molar basis) for the combustion of n-octane at $25^{\circ}$ C would be: