For a chemical reaction A rightarrow B,it was | vedclass

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(D) The reaction is $A \rightarrow B$.Given that the concentration of $B$ increases by $\Delta[B] = 0.2 \, mol \, L^{-1}$ in time $\Delta t = 30 \, mi

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) The reaction is $A ightarrow B$.Given that the concentration of $B$ increases by $\Delta[B] = 0.2 \, mol \, L^{-1}$ in time $\Delta t = 30 \, min$.Convert time to hours: $\Delta t = 30 \, min = 0.5 \, h$.The average rate of reaction is given by $	ext{Rate} = \frac{\Delta[B]}{\Delta t}$.$	ext{Rate} = \frac{0.2 \, mol \, L^{-1}}{0.5 \, h} = 0.4 \, mol \, L^{-1} \, h^{-1}$.We need to express this as $...... 	imes 10^{-1} \, mol \, L^{-1} \, h^{-1}$.$0.4 = 4 	imes 10^{-1}$.Thus,the value is $4$.

$t \ (s)$	$0$	$30$	$60$	$90$
$[\text{Ester}] \ (mol \ L^{-1})$	$0.55$	$0.31$	$0.17$	$0.085$

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)

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