For the reaction- A g -2 B g - 2 C g -3 D g The value of - H at 27- C is 19-0 k cal- The value of - E for the reaction would beA- 19-8 kcalB- 18-8 kcalC- 17-8 kcalD- 20-8 kcal

The correct option is C 17.8 kcalGiven reaction is, A(g) + 2B(g)→ 2C(g) + 3D(g) ∴Δ n (g) nbsp;= nbsp; 5 3 = 2 We know, Δ H = Δ E + Δ n_g RT ⇒Δ E = Δ H ...

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Standard XII

Chemistry

Difference between Internal Energy and Enthalpy

For the react...

Question

For the reaction,
$A (g) + 2 B (g) \to 2 C (g) + 3 D (g)$
The value of $Δ H$ at $27^{\circ} C$ is $19.0 k c a l$ . The value of $Δ E$ for the reaction would be

20.8 k c a l

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19.8 k c a l

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18.8 k c a l

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17.8 k c a l

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Solution

The correct option is D $17.8 k c a l$
Given reaction is,
$A (g) + 2 B (g) \to 2 C (g) + 3 D (g)$
$∴ Δ n (g) = 5 - 3 = 2$
We know,
$Δ H = Δ E + Δ n_{g} R T \Rightarrow Δ E = Δ H - Δ n_{g} R T = 19 - (2 \times 2 \times 10^{- 3} \times 300) = 19 - 1.2 k c a l = 17.8 k c a l .$

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For the reaction, A(g)+2B(g)→2C(g)+3D(g) The value of ΔH at 27 ∘C is 19.0 k cal. The value of ΔE for the reaction would be

The correct option is D 17.8 kcalGiven reaction is, A(g)+2B(g)→2C(g)+3D(g) ∴Δn(g)=5−3=2 We know, ΔH=ΔE+ΔngRT⇒ΔE=ΔH−ΔngRT =19−(2×2×10−3×300) =19−1.2 kcal=17.8 kcal.

For the reaction,
$A (g) + 2 B (g) \to 2 C (g) + 3 D (g)$
The value of $Δ H$ at $27^{\circ} C$ is $19.0 k c a l$ . The value of $Δ E$ for the reaction would be

The correct option is D $17.8 k c a l$
Given reaction is,
$A (g) + 2 B (g) \to 2 C (g) + 3 D (g)$
$∴ Δ n (g) = 5 - 3 = 2$
We know,
$Δ H = Δ E + Δ n_{g} R T \Rightarrow Δ E = Δ H - Δ n_{g} R T = 19 - (2 \times 2 \times 10^{- 3} \times 300) = 19 - 1.2 k c a l = 17.8 k c a l .$