In the reaction : P + Q longrightarrow | vedclass

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Question

For reactant '$P$'

since $t_{75\%}=2\, t_{50\%}$

therefore order with respect to '$P$' is 'one' For reactant '$Q$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

For reactant '$P$'

since $t_{75\%}=2\, t_{50\%}$

therefore order with respect to '$P$' is 'one' For reactant '$Q$'

linear variation of concentration of '$Q$' with time suggest that order with respect to '$Q$' is zero therefore overall order will be $=1+0=1$

Correct Option $(D)$

(Assuming conc. of other reactant to be very large)

Exp.	$[A]$	$[B_2]$	Rate $(mol\,L^{-1}\,S^{-1})$
$1$	$0.50$	$0.50$	$1.6 \times {10^{ - 4}}$
$2$	$0.50$	$1.00$	$3.2 \times {10^{ - 4}}$
$3$	$1.00$	$1.00$	$3.2 \times {10^{ - 4}}$

${N_2}$ in $cc$	$6.25$	$9.50$	$11.42$	$13.65$	$35.05$
Time (minutes)	$10$	$15$	$20$	$25$	Finally

In the reaction : $P + Q \longrightarrow R + S$ the time taken for $75\%$ reaction of $P$ is twice the time taken for $50\%$ reaction of $P$. The concentration of $Q$ varies with reaction time as shown in the figure. The overall order of the reaction is

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