Question

In the reaction A+3B gives 2C the rate of formation of C is

Answer 1

Answer:

A+2B---2C

Rate law,

-1/2dC/dt=dA/dt+1/2dB/dt

rate of formation Of C is,

-dC/dt=2dA/dt+dB/dt

Answer 2

For the reaction $A + 2B \rightarrow 2C$, the rate of formation of C is $\frac{-dC}{dt} = \frac{2dA}{dt} + \frac{dB}{dt}$.

Explanation:

Rate law is defined as the function of concentration of reactants in a chemical reaction. Rate law of a reaction depends on the slow step of a chemical reaction.

For example, for the reaction $A + 2B \rightarrow 2C$

The rate law for this reaction is as follows.

       $\frac{-1}{2}\frac{dC}{dt} = \frac{dA}{dt} + \frac{1}{2}\frac{dB}{dt}$

Hence, the rate of formation of C will be given as follows.

      $\frac{-dC}{dt} = \frac{2dA}{dt} + \frac{dB}{dt}$

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