For reaction : A ( g ) ⇌ 2 D ( g ) + 3 C ( g ) , Find the difference in ( ln K P ∘ ) at 500 K and ( ln K P ∘ ) at 300 K . (R = 8 J/mole-K)

Question

For reaction : A ( g ) ⇌ 2 D ( g ) + 3 C ( g ) , Find the difference in ( ln K P ​ ∘ ) at 500 K and ( ln K P ​ ∘ ) at 300 K . (R = 8 J/mole-K)

TestHub · Accepted Answer

Explain Question:

We have to find the difference of $ln Kp^{\circ}$ at 500 k and $ln Kp^{\circ}$ at 300 K .

Concept: We have to find the overall $Δ H$ value of by combining the given equations then use Van't Hoff equation.

Solution:

Given:

A(g) ⇌ 2B(g) + C(g) ; $Δ H_{1} = 50 k J / m o l$

B(g) ⇌ D(g) + C(g) ; $Δ H_{2} = 35 k J / m o l$

A(g) ⇌ 2B(g) + C(g) ; $Δ H_{1} = 50 k J / m o l$

2[B(g) ⇌ D(g) + C(g)] ; $Δ H_{2} = 2 \times 35 k J / m o l = 70 k J / m o l$

Overall reaction

A(g) ⇌ 2D(g) + 3C(g) ; $Δ H = 50 k J / m o l + 70 k J / m o l = 120 k J / m o l$

Using the Van't Hoff equation:

$ln K_{2} - ln K_{1} = \frac{Δ H}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

$= \frac{120 \times 1000}{8} (\frac{1}{300} - \frac{1}{500}) = 20$

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