The value of enthalpy change $Δ H$ for the reaction

$C_{2} H_{5} O H (1) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (1) a t 27^{\circ} C i s - 1366.5 k J m o l^{- 1},$ The value of internal

energy change for the above reaction at this temperature will be:

-1369 kJ

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-1364.0 kJ

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-1361.5 kJ

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-1371.5 kJ

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Solution

The correct option is B
-1364.0 kJ

$C_{2} H_{5} O H (1) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (1) Δ n_{g} = 2 - 3 = - 1 Δ U = Δ H = Δ n_{g} R T = - 1366.5 - (1) \times \frac{8.314}{10^{3}} \times 300 = - 1366.5 + 0.8314 \times 3 = - 1364 k J$

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Similar questions

Δ H

C_{2} H_{5} O H_{(l)} + 3 O_{2 (g)} \to 2 C O_{2 (g)} + 3 H_{2} O_{(l)}

27^{0}

- 1366.5 k J m o l^{- 1}

(Δ H)

C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)

27^{0} C

m o l^{- 1}

The enthalpy change for the combustion of ethanol at 298 K and 1 atm is given below

CH₃-CH₂OH(l) + 3O₂(g)------------à 2CO₂ (g)+ 3H₂O(l) ∆H=-1363Kj.Calculate the internal energy change for the above reaction?

C_{2} H_{4} (g) + 3 O_{2} (g) ⟶ 2 C O_{2} (g) + 2 H_{2} O (g)

C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 {C O}_{2} (g) + 2 H_{2} O (g)

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