For the hypothetical reaction A 2 g + B 2 g ⇌ 2 AB g Δ r G ∘ and Δ r S ∘ are 20 kJ/mol and - 20 JK -1 mol -1 respectively at 200 K. If Δ r C p is 20 JK -1 mol -1 then Δ r H ∘ at 400 K is :

Δ G 2 0 0 ∘ = Δ H 2 0 0 ∘ - T Δ S 2 0 0 ∘ Δ H 2 0 0 ∘ = 2 0 - 4 = 16 kJ/mol Δ H T 2 ∘ = Δ H T 1 ∘ + Δ C P T 2 - T 1 Δ H 4 0 0 ∘ = Δ H 2 0 0 ∘ + 2 0 × 2 0 0 1 0 0 0 kJ/mol = 1 6 + 4 = 20 kJ/mol

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Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IFirst law of ThermodynamicsEasy2 minPYQ_2020

ChemistryEasysingle choice

For the hypothetical reaction

$A_{2} (g) + B_{2} (g) ⇌ 2 AB (g)$

$Δ_{r} G^{\circ}$ and $Δ_{r} S^{\circ}$ are 20 kJ/mol and $- {20 JK}^{-1} {mol}^{-1}$ respectively at 200 K.

If $Δ_{r} C_{p}$ is ${20 JK}^{-1} {mol}^{-1}$ then $Δ_{r} H^{\circ}$ at 400 K is :

Options:

Answer:

Solution:

$Δ G_{200}^{\circ} = Δ H_{200}^{\circ} - T Δ S_{200}^{\circ}$

$Δ H_{200}^{\circ} = 20 - 4 = 16 kJ/mol$

$Δ H_{T_{2}}^{\circ} = Δ H_{T_{1}}^{\circ} + Δ C_{P} [T_{2} - T_{1}]$

$Δ H_{400}^{\circ} = Δ H_{200}^{\circ} + \frac{20 \times 200}{1000} kJ/mol$

$= 16 + 4 = 20 kJ/mol$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics

⏱ 2mℹ️ Source: PYQ_2020

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