Question

For the reaction H₂(g)+Br₂(g)—>2HBr(g). write the equation for reaction rate.

Answer 1

please see the attached image this is the rate law expression for the following reaction

Answer 2

Answer:

$\dfrac{-dH_{2} }{dt} = \dfrac{-dBr_{2} }{dt} = \dfrac{+dHBr }{2dt}$

Explanation:

We have given the reaction as  H₂(g)+Br₂(g)—>2HBr(g).

Rate of reaction is defined as the speed at which reactants are decreased and product are increased at a same time.

For a same reaction reactant decreased while product increased, but after reaching at a some point there is no conversion of reactant to product,at that time rate of reaction becomes zero.

The rate of reaction for the given reaction is

$\dfrac{-dH_{2} }{dt} = \dfrac{-dBr_{2} }{dt} = \dfrac{+dHBr }{2dt}$