Question

Explain yaar
dont give direct answer. .....Plz i beg u​

Answer 1

Answer: option a

Explanation: sure binod xD

Answer 2

For the given reaction ,

$\displaystyle \mathtt{2NH_3 \rightarrow N_2 + 3H_2}$

we know that ,

rate of dissociation of reactant = rate of formation of product

Hence , the rate of the reaction is written as ,

$\boxed{ \pink{ \displaystyle \mathtt{ - \frac{1}{2} \frac{d(NH_3)}{dt} = + \frac{d(N_2)}{dt} = + \frac{1}{3} \frac{d(H_2)}{dt}}}}$

Note :

• negative sign denotes decrease in the concentration of reactant

• positive sign denotes increase in the concentration of product

Given reaction

$\displaystyle \star - \mathtt{ \frac{d(NH_3)}{dt} = k_1(NH_3)}$

$\displaystyle \star\mathtt{ \frac{d(N_2)}{dt} = k_2(NH_3)}$

$\displaystyle \star\mathtt{ \frac{d(H_2)}{dt} = k_3(NH_3)}$

Solution:

Substituting the above values in the rate equation we get ,

$\rightarrow\displaystyle \mathtt{ \frac{1}{2}k_1 (NH_3) = +k_2 (NH_3)= + \frac{1}{3} k_3 (NH_3)}$

Multiplying the entire equation by 3 we get,

$\rightarrow\displaystyle \mathtt{ \frac{3}{2}k_1 (NH_3) = +3k_2 (NH_3)= + \frac{3}{3} k_3 (NH_3)}$

$\boxed {\red{\displaystyle \mathtt{ 1.5k_1 = +3k_2 = + k_3 }}}$