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In an experiment (reactant $\rightarrow$ product) we observe that if we increase the reactant to twice then time taken in the concentration of reaction to drop to $\large\frac{1}{2}$ of its value is reduced to $\large\frac{1}{4}$ then order is

$(a)\;0\qquad(b)\;1\qquad(c)\;2\qquad(d)\;3$

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1 Answer

$t_{\large\frac{1}{2}}\propto [A]^{1-n}$
$t_{\large\frac{1}{2}}=P [A]^{1-n}$
A = Rate of concentration of the reaction
t = Time taken
P=constant of proportionality
$\large\frac{1}{4}$$t_{\large\frac{1}{2}}=P[2A]^{1-n}$
$4=[\large\frac{1}{2}]^{1-n}$
$4=[2]^{n-1}$
$2^2=[2]^{n-1}$
$n=3$
Hence (d) is the correct answer.
answered Dec 13, 2013 by sreemathi.v
edited Dec 27, 2014 by sharmaaparna1
 

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