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For the reversible reaction $N_{2} + O_{2} ⇌ 2 N O$ activation energy of the forward reaction is greater than that of backward reaction. The slope of In K Vs I/T (abscissa) graph will be:

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$\frac{Δ H}{2.303 R}$

$- \frac{Δ H}{2.303 R}$

$- \frac{Δ H}{R}$

Zero

answer is D.

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Detailed Solution

In K = Constant $- \frac{Δ H}{R T}$

Since, $E_{a} (forward) > E_{a} (backward),the reaction is$

Endothermic, $Δ H > 0, Hence slope = - \frac{Δ H}{R}$

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