Question

For the chemical reaction ${{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)$ the correct option is
(A) $-\dfrac{1}{3}\dfrac{d[{{H}_{2}}]}{dt}=-\dfrac{1}{2}\dfrac{d[N{{H}_{3}}]}{dt}$
(B) $-\dfrac{d[{{N}_{2}}]}{dt}=2\dfrac{d[N{{H}_{3}}]}{dt}$
(C) $-\dfrac{d[{{N}_{2}}]}{dt}=\dfrac{1}{2}\dfrac{d[N{{H}_{3}}]}{dt}$
(D) $3\dfrac{d[{{H}_{2}}]}{dt}=2\dfrac{d[N{{H}_{3}}]}{dt}$

Answer 1

As the reaction proceeds, the number of reactants decreases and the number of products increase in a chemical reaction. The number of reactants consumed or the number of products formed depends on the overall rate of the reaction. The rate of the formation or the rate of disappearance is equal to the overall rate of the reaction.

Complete step by step solution:
For a chemical reaction, the measure of the change in concentration of reactants or the change in concentration of products per unit time is known as the reaction rate.
Consider the stoichiometrically complicated reaction,
$xX+yY\to aA+bB$
The rate of reaction = rate of disappearance of reactant = rate of formation of products
$r =$ $-\dfrac{1}{x}\dfrac{d[X]}{dt}=-\dfrac{1}{y}\dfrac{d[Y]}{dt}=\dfrac{1}{a}\dfrac{d[A]}{dt}=\dfrac{1}{b}\dfrac{d[B]}{dt}$ -- (1)
Where r = rate of overall reaction
$d[X],d[Y],d[A],\And d[B]$, are the change in the concentrations of reactants and products, and $dt$ will be changed in time.
Apply this equation (1) for the given reaction
${{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)$
Rate of the reaction = rate of disappearance of ${{N}_{2}}(g)$ and ${{H}_{2}}(g)$ = rate of formation of $N{{H}_{3}}(g)$
Then the rate of reaction is given as,
$-\dfrac{d[{{N}_{2}}]}{dt}=-\dfrac{1}{3}\dfrac{d[{{H}_{2}}]}{dt}=\dfrac{1}{2}\dfrac{d[N{{H}_{3}}]}{dt}$

Hence, the correct answer is option C.

Note: Based on the collision theory, reactant molecules collide with each other to form products. The number of colliding particles will increase, then the rate of reaction increases by increasing the concentration of reaction.