Internal energy change for combustion of ethanol is 416.5 $k J m o l^{- 1}$ . The calorific value of ethanol at 300 K in $(k J)$ is (Take R= 25/3 J/molK)

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Solution

$C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)$
$Δ H = Δ U + Δ n_{g} R T$
$Δ H = 416.5 - \frac{R T}{1000}$
$Δ H = 416.5 - \frac{25 \times 300}{3 \times 1000} = 414 k J$
Calorific value = $\frac{414}{46} = 9 k J$

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Internal energy change for combustion of ethanol is 416.5 kJ mol−1. The calorific value of ethanol at 300 K in (kJ) is (Take R= 25/3 J/molK)

C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l) ΔH=ΔU+ΔngRT ΔH=416.5−RT1000 ΔH=416.5−25×3003×1000=414 kJ Calorific value = 41446=9 kJ

Internal energy change for combustion of ethanol is 416.5 $k J m o l^{- 1}$ . The calorific value of ethanol at 300 K in $(k J)$ is (Take R= 25/3 J/molK)

$C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)$
$Δ H = Δ U + Δ n_{g} R T$
$Δ H = 416.5 - \frac{R T}{1000}$
$Δ H = 416.5 - \frac{25 \times 300}{3 \times 1000} = 414 k J$
Calorific value = $\frac{414}{46} = 9 k J$