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At a certain temperature the half change periof for the catalystic decompoistion of ammonia was found as follows: `|{:("Pressure (Pa)",6667,13333,26666),("Half life periof in hours",3.52,1.92,1.0):}|` Calculate the order of reaction. |
Answer» `((t_(1//2))_(1))/((t_(1//2))_(2))=((a_(2))/(a_(1)))^(n-1)` where, n is order of reaction From the given data, `(3.25)/(1.92)=((13333)/(6667))^(n-1)" " (a prop"initial pressure")` `log .(3.52)/(1.92)=(n-1)log2=0.3010xx(n-1)` `0.2632=0.30120xx(n-1)` `n=1.87~~2` Similar calculations are made between first and third observation. n comes equal to `1.908(~~2)` Thus, the reaction is of second. order. |
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