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  • For a reaction, A + 2B  products. r = k [A]0[B]0 Initial concentra

For a reaction, A + 2B \rightarrow products. r = k [A]0[B]0
Initial concentration of B = B0. Then the half life of B will be? 

Option: 1

\frac{\mathrm B_{0}}{4\mathrm k}


Option: 2

\frac{\mathrm B_{0}}{2\mathrm k}


Option: 3

\frac{\mathrm B_{0}}{\mathrm k}


Option: 4

{\mathrm B_{0}


Answers (1)

best_answer

Given reaction:

\mathrm{A+2B \rightarrow products}

Rate (r) = k [A]0[B]0

It will be a zero-order reaction due to rate is not depending on reactant concentration.

According to the question, we have given:

\mathrm{r\: =\: k[A]^{0}[B]^0\: =\: -\frac{1}{2}\frac{dB}{dt}}

\mathrm{-\frac{1}{2}\frac{dB}{dt}\: =\: k\quad\quad\quad(Since\: [A]^{0}\: =\: 1 \ and\ [B]^{0}\: =\: 1 )}

Now, integrating both sides we get:

\mathrm{[B]^{B}_{B_{o}}\: =\: -2k(t)^{t}_{o}}

\mathrm{B-B_{o}\: =\: -2kt}

\mathrm{\mathbf{B\: =\: B_{o}-2kt}}
This is the concentration of A after time 't'.

Now, after half-life of a reaction :

\mathrm{\mathbf{B\: =\: \frac{B_{o}}{2}}}

So, 

\mathrm{\mathbf{\frac{B_o}{2}\: =\: B_{o}-2kt_{(1/2)}}}

\mathrm{t_{1/2}\: =\: \frac{B_{o}}{2(2k)} }

\mathrm{Thus,\: \mathbf{t_{1/2}\:=\: \frac{B_{o}}{4k}}}
This is the half-life of B for this reaction.

Therefore, option(1) is correct

Posted by

shivangi.shekhar

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