Question

in a certain reaction shown below,
4A + 2B gives 3C

If rate of formation of C is 9.6 10-2 mol/Ls, what will be rate of reaction?
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Answer 1

Answer : The rate of reaction is, $3.2\times 10^{-2}moleL^{-1}s^{-1}$

Explanation : Given,

Rate of formation of C = $9.6\times 10^{-2}moleL^{-1}s^{-1}$

The given rate of reaction is,

$4A+2B\rightarrow 3C$

The expression for rate of reaction :

$\text{Rate of disappearance}=-\frac{1}{4}\frac{d[A]}{dt}=-\frac{1}{2}\frac{d[B]}{dt}$

$\text{Rate of formation}=\frac{1}{3}\frac{d[C]}{dt}$

Now we have to calculate the rate of reaction.

$\text{Rate of reaction}=\frac{1}{3}\frac{d[C]}{dt}=\frac{1}{3}\times 9.6\times 10^{-2}moleL^{-1}s^{-1}=3.2\times 10^{-2}moleL^{-1}s^{-1}$

Therefore, the rate of reaction is, $3.2\times 10^{-2}moleL^{-1}s^{-1}$

Answer 2

The rate of the reaction is $3.2\times 10^{-2}Ms^{-1}$

Explanation:

We are given:

Rate of formation of C = $\frac{d[C]}{dt}=9.6\times 10^{-2}Ms^{-1}$

For the given chemical equation:

$4A+2B\rightarrow 3C$

As we know, rate of the reaction remains the same.

$\text{Rate of the reaction}=\frac{1}{3}\frac{d[C]}{dt}$

Putting values in above equation, we get:

$\text{Rate of the reaction}=\frac{1}{3}\times 9.6\times 10^{-2}Ms^{-1}\\\\\text{Rate of the reaction}=3.2\times 10^{-2}Ms^{-1}$

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