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

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Solution :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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