For the hypothetical reaction- A2g-B2 g - 2AB g If - rG- and - rS- are 20 kj-mol and -20 JK-1mol-1 respectively at 200 K - rCP is 20JK-1mol-1 then value - rH-10 of KJ-mol at 400 K is-

Δ H^o_r (200)=Δ G^o_r (200)+TΔ S^o_r (200)=20000 J/mol + 200 K × ( 20 J/K/mol)=16000 J/mol = 16 kJ/mol Δ H^o_r (400)=Δ H^o_r (200)+Δ _r C_P ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

For the hypot...

Question

For the hypothetical reaction,
$A_{2} g + B_{2} (g) ⇌ 2 A B (g)$
If $Δ_{r} G^{\circ} a n d Δ_{r} S^{\circ}$ are 20 kj/mol and -20 $J K^{- 1} m o l^{- 1}$ respectively at 200 K
$Δ_{r} C_{P} i s 20 J K^{- 1} m o l^{- 1}$ then value $\frac{Δ_{r} H^{\circ}}{10}$ of (KJ/mol) at 400 K is:

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Solution

$Δ H_{r}^{o} (200) = Δ G_{r}^{o} (200) + T Δ S_{r}^{o} (200) = 20000 J / m o l + 200 K \times (- 20 J / K / m o l) = 16000 J / m o l = 16 k J / m o l$
$Δ H_{r}^{o} (400) = Δ H_{r}^{o} (200) + Δ_{r} C_{P} (T_{2} - T_{1}) = 16 k J / m o l + \frac{20 J / m o l (400 K - 200 K)}{1000} = 16 : k J / m o l + 4 : k J / m o l = 20 : k J / m o l$
$\frac{Δ H_{r}^{o} (400)}{10} = \frac{20 : k J / m o l}{10} = 2 : k J / m o l$

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For the hypothetical reaction,A2g+B2(g)⇌2AB(g)If ΔrG∘andΔrS∘ are 20 kj/mol and -20 JK−1mol−1 respectively at 200 KΔrCPis20JK−1mol−1 then value ΔrH∘10 of (KJ/mol) at 400 K is:

ΔHor(200)=ΔGor(200)+TΔSor(200)=20000J/mol+200K×(−20J/K/mol)=16000J/mol=16kJ/molΔHor(400)=ΔHor(200)+ΔrCP(T2−T1)=16kJ/mol+20J/mol(400K−200K)1000=16 :kJ/mol+4 :kJ/mol=20 :kJ/molΔHor(400)10=20 :kJ/mol10=2 :kJ/mol

For the hypothetical reaction,
$A_{2} g + B_{2} (g) ⇌ 2 A B (g)$
If $Δ_{r} G^{\circ} a n d Δ_{r} S^{\circ}$ are 20 kj/mol and -20 $J K^{- 1} m o l^{- 1}$ respectively at 200 K
$Δ_{r} C_{P} i s 20 J K^{- 1} m o l^{- 1}$ then value $\frac{Δ_{r} H^{\circ}}{10}$ of (KJ/mol) at 400 K is: