1.

Calculate the order of reaction for the reaction `2NH_(3)(g) to N_(2)(g) + 3H_(2)(g)` Given that half life period `(t_(1//2))` under a pressure of 50mm Hg is 3.52 and under a pressure of 100 mm Hg is 1.82

Answer» We know that
`n=1+(log(t_(1//2))1-log(t_(1//2))2)/(log p_(2)-log p_(1))`
`1+ (log 3.52 - log 1.82)/(log 100 - log 50)`
`1+(0.5465 - 0.2600)/(2-1.6989)`
`1+ (0.2865)/(0.3011) = 1+0.9515=2`


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