For the reaction, $S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {S O}_{3} (g)$ , $K_{p}$ is $32 a t m^{- 1 / 2}$ at $800 K$ and $Δ H$ for the reaction is $- 187.9 k J$ ${m o l}^{- 1}$ . Calculate its value at $900 K$ , if it is assumed that $Δ H$ remains constant over this range of temperature.

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Solution

The given reaction is :-
${S O}_{2} + \frac{1}{2} O_{2} ⇌ {S O}_{3} (g)$
$K_{P} = 32 {a t m}^{- 1 / 2} a t 800 K$
$△ H = - 187.9 K J / m o l$
By Vant' Hoff equation :-
$l o g (\frac{{(K_{P})}_{2}}{{(K_{P})}_{1}}) = \frac{△ H}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
$\Rightarrow l o g ({K_{P}}_{2}) - l o g ({K_{P}}_{1}) = \frac{- 187.9 \times 1000}{2.303 \times 8.314} (\frac{900 - 800}{900 \times 800})$
$\Rightarrow l o g {K_{P}}_{2} = \frac{- 187.9 \times 1000 \times 100}{2.303 \times 8.314 \times 900 \times 800} + l o g (32)$
$\Rightarrow l o g {K_{P}}_{2} = 1.36 + 1.50$
$\Rightarrow K_{P_{2}} = A n t i l o g (2.86) = 724.44$

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For the reaction, SO2(g)+12O2(g)⇌SO3(g), Kp is 32 atm−1/2 at 800K and ΔH for the reaction is −187.9kJ mol−1. Calculate its value at 900K, if it is assumed that ΔH remains constant over this range of temperature.

For the reaction, $S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {S O}_{3} (g)$ , $K_{p}$ is $32 a t m^{- 1 / 2}$ at $800 K$ and $Δ H$ for the reaction is $- 187.9 k J$ ${m o l}^{- 1}$ . Calculate its value at $900 K$ , if it is assumed that $Δ H$ remains constant over this range of temperature.