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  •  Consider an single step reaction as follows: <img alt="\mathrm{\mathrm{A}+213 \longrightarrow \mathrm{C}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cmathrm%7B

 Consider an single step reaction as follows:
\mathrm{\mathrm{A}+213 \longrightarrow \mathrm{C}},
If the rate constant \mathrm{\left ( k \right )} for the reaction is \mathrm{0.02 \cdot L^{2} \mathrm{~mol}^{-2} \mathrm{~S}^{-1}},  what will be the rate of disseppearence of B. If the initial concentration of \mathrm{A \& B} are \mathrm{0.1\, mol\, L^{-1}} ?

Option: 1

\mathrm{2 \times 10^{-5} L^{2} mol ^{-2} s^{-1}}


Option: 2

\mathrm{1 \times 10^{4} L^{2} mol ^{-2} s^{-1}}


Option: 3

\mathrm{4 \times 10^{-5} L^{2} mol ^{-2} s^{-1}}


Option: 4

\mathrm{8 \times 10^{4} L^{2} mol ^{-2} s^{-1}}


Answers (1)

best_answer

Rate law equation for the given elementary reaction is as follow

\mathrm{\text { Rate }=K[A] [B]^{2}}
\mathrm{\text { Rate }=0.02 \times 0.1 \times(0.1)^{2}}
            \mathrm{=2 \times 10^{-5} L^{2} \text { mol }^{-2} s^{-1}}

\mathrm{ \text { Rate }=-\frac{1}{2}\left[\frac{d(B)}{d t}\right] }

              \mathrm{ -\frac{d(B)}{d t}=2 \times \text { Rate }=2 \times 2 \times 10^{-5} }
Rete of disappearance of  \mathrm{ B=\frac{-d B}{d t}=4 \times 10^{-5} \mathrm{~L}^{2} \mathrm{mol}^{-2} \mathrm{~s}^{-1}}

Posted by

Rakesh

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