A) Order of reaction is 1.5
B) Molecularity of the reaction is 2
C) By increasing the concentration of \[B{{r}_{2}}\] four times the; rate of reaction is doubled
D) All the above are correct
Correct Answer: D
Solution :
The given reaction \[{{H}_{2}}+B{{r}_{2}}\xrightarrow{{}}2HBr\] is a bi-molecular reaction as moles of reactant are 2. The order can be determined as: \[\frac{dx}{dt}=k[{{H}_{2}}]{{[B{{r}_{2}}]}^{1/2}}\] Order \[=\frac{1+1}{2}=\frac{3}{2}=1.5\] If conc. of \[B{{r}_{2}}\] is increased 4 times rate \[k[{{H}_{2}}]{{[4B{{r}_{2}}]}^{1/2}}\] \[=2k[{{H}_{2}}]{{[B{{r}_{2}}]}^{1/2}}\] \[=2\times \operatorname{inttial}\,rate\] i.e., rate increases 2 times. Hence, all the statements are correct
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