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Question

(A) The rate of reaction is given by the expression: $	ext{Rate} = -\frac{1}{4} \frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt} = \frac{1}{6} \frac{d[

Vedclass · Accepted Answer

$1 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$ and $30 	imes 10^{-2} \ mol \ L^{-1}$

Vedclass · Answer

(A) The rate of reaction is given by the expression: $	ext{Rate} = -\frac{1}{4} \frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt} = \frac{1}{6} \frac{d[C]}{dt} = \frac{1}{9} \frac{d[D]}{dt}$.Given,rate of disappearance of $A$ $(-\frac{d[A]}{dt})$ $= 4 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$.Therefore,$	ext{Rate of reaction} = \frac{1}{4} 	imes (4 	imes 10^{-2}) = 1 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$.Now,$	ext{Rate of disappearance of } B = 3 	imes 	ext{Rate of reaction} = 3 	imes 1 	imes 10^{-2} = 3 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$.Amount of $B$ consumed in $10 \ s = 	ext{Rate of disappearance of } B 	imes 	ext{Time} = (3 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}) 	imes 10 \ s = 30 	imes 10^{-2} \ mol \ L^{-1}$.

Time $t$ $(s)$	$0$	$5$	$10$	$20$	$30$
Concentration $(mol \ L^{-1})$	$160 \times 10^{-3}$	$80 \times 10^{-3}$	$40 \times 10^{-3}$	$10 \times 10^{-3}$	$3.5 \times 10^{-3}$

Similar Questions

For the reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ under certain conditions of temperature and partial pressure of the reactants,the rate of formation of $NH_3$ is $0.001 \ kg \ L^{-1} \ h^{-1}$. The rate of consumption of $H_2$ under the same condition is $..... \ kg \ L^{-1} \ h^{-1}$.

For the reaction $N_2 + 3H_2 \to 2NH_3$,which of the following represents the correct expression for $d[NH_3]/dt$?

For a chemical reaction $A \rightarrow B$,it was found that the concentration of $B$ increases by $0.2 \, mol \, L^{-1}$ in $30 \, min$. The average rate of the reaction is $...... \times 10^{-1} \, mol \, L^{-1} \, h^{-1}$. (Nearest integer)

Which one of the following equations is correct for the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$?

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