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Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant

For the disso...

Question

For the dissociation of $P C l_{5} (g)$
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
Slope of the linear curve is such that $θ = t a n^{- 1} (- 2.1)$

Thus, $△ H^{o}$ is

28.72 J m o l^{- 1}

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- 28.72 J m o l^{- 1}

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40.2 J m o l^{- 1}

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- 12.47 J m o l^{- 1}

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Solution

The correct option is C $40.2 J m o l^{- 1}$
Van't Hoff equation is
$l o g_{10} K = - \frac{△ H^{o}}{2.303 R T} + l o g C$
where C is a constant.
$Graph between l o g_{10} K a n d \frac{1}{T} is linear .$
On comparing it to
$y = m x + c$
slope, $m = - \frac{△ H^{o}}{2.303 R} . . . . .1$
but, $slope m also equals, t a n θ$
since $θ = t a n^{- 1} (- 2.1)$
So,
$s l o p e, m = t a n θ = - 2.1 . . . . . .2$
comparing equation 1 and 2 , we get
$- \frac{△ H^{o}}{2.303 R} = t a n θ$

putting the values we get,
$∴ △ H^{o} = 2.303 R \times 2.1$
$= 2.303 \times 8.314 \times 2.1 = 40.2 J m o l^{- 1} .$

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[DCE 2000; AIEEE 2005]

For the reaction, $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ , the degree of dissociation 'x' is related to pressure 'P

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