-For the reaction C2H4 g- 3O2 g- 2CO2 g - 2H2 Ol- - E - -1415kJ- The -H at 27 0C is

The correct option is B 1420 kJ Δ H=Δ E+Δ n_gRT = 1415 + ( 2) × 8.314 × 300 × 10 ^ 3 = 1415 4.98 kJ Δ n_g= ...

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Standard XII

Chemistry

Enthalpy of Combustion

:For the reac...

Question

:For the reaction
$C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l);$
$Δ$ $E$ = $- 1415 k J$ . The $Δ$ H at 27 $^{0}$ C is

- 1410 k J

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- 1420 k J

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+ 1420 k J

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+ 1410 k J

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Solution

The correct option is B $- 1420 k J$
$Δ H = Δ E + Δ n_{g} R T$
$=$ $- 1415 + (- 2) \times 8.314 \times 300 \times$ 10 $^{- 3}$
$=$ $- 1415 - 4.98 k J$
$Δ n_{g} =$ $2 - 3 - 1$
$=$ $- 2$
$≃$ $- 1419.98$
$≃$ $- 1420 k J$

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Similar questions

Q. For reaction

C_{2} H_{4} (g) + 3 O_{2} (g) ⟶ 2 C O_{2} (g) + 2 H_{2} O (l); Δ U = - 1415 k J

. Then

Δ H

27^{o} C

is :

Q. For the reaction:

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

Δ E = - 1415 K J

. The

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If $Δ$ E is the heat of reaction for $C_{2} H_{5} O H_{(l)} + 3 O_{2 (g)} \to 2 C O_{2 (g)} + 3 H_{2} O_{(l)}$ at constant volume, the $Δ$ H (Heat of reaction at constant pressure) at constant temperature is:

Q. For the reaction,

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

;

Δ U = - 1415

kJ. Then

Δ H

27^{o}

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For the reaction $C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l)$ , the difference between enthalpy change and internal energy change is:

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:For the reactionC2H4(g)+3O2(g)→2CO2(g)+2H2O(l);ΔE = −1415kJ. The ΔH at 270C is

The correct option is B −1420kJΔH=ΔE+ΔngRT= −1415+(−2)×8.314×300× 10−3= −1415−4.98kJΔng= 2−3−1= −2≃ −1419.98≃ −1420kJ

:For the reaction
$C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l);$
$Δ$ $E$ = $- 1415 k J$ . The $Δ$ H at 27 $^{0}$ C is

The correct option is B $- 1420 k J$
$Δ H = Δ E + Δ n_{g} R T$
$=$ $- 1415 + (- 2) \times 8.314 \times 300 \times$ 10 $^{- 3}$
$=$ $- 1415 - 4.98 k J$
$Δ n_{g} =$ $2 - 3 - 1$
$=$ $- 2$
$≃$ $- 1419.98$
$≃$ $- 1420 k J$