In a reaction, $2 \mathrm{~A} ightarrow$ products, the concentration of $A$ decreases from $0.50 \mathrm{M}$ to $0.38 \mathrm{M}$ in $10 \mathrm{~min}$. What is the rate of disappearance of A (in $\mathrm{M} \mathrm{s}^{-1}$ ) during this interval?

Question

Aicharya · Accepted Answer

R.O.D of A $=\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}$
Given, $[\mathrm{A}]_{\text {initial }}=0.50 \mathrm{M}$
$[\mathrm{A}]_{\text {final }}=0.38 \mathrm{M}$
$\mathrm{dt}=10 \mathrm{~min}=600 \mathrm{sec}$
$\mathrm{d}[\mathrm{A}]=0.12$
Rate $=\frac{0.12}{600}=2 \times 10^{-4} \mathrm{s}^{-1}$.

In a reaction, $2 \mathrm{~A} ightarrow$ products, the concentration of $A$ decreases from $0.50 \mathrm{M}$…

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