Question

Give the expressions for the disappearance of the reactants first, in the order written in the chemical equation. Then write the expressions for the appearance of the products in the order written in the chemical equation.

Write the expressions in order of appearance in the equation in the form.

± 1/X . Δ[A]/ Δt

where ± is either a plus OR a minus sign, not both,
X is an integer, and
A is a chemical species.

4NH3(g)+ 5O2(g) → 4NO2+ 6H2O(g)

Answer 1

Answer: Rate of disappearance of $NH_3$ = $\frac{-\Delta [NH_3]}{4dt}]$

Rate of disappearance of $O_2$ = $\frac{-\Delta [O_2]}{5dt}$

Rate of appearance of $NO_2$ = $\frac{+\Delta [NO_2]}{4dt}$

Rate of appearance of $H_2O$ = $\frac{+\Delta [H_2O]}{6dt}$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

For the given chemical equation :

$4NH_3(g)+5O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)$

Rate of disappearance of $NH_3$ = $\frac{-\Delta [NH_3]}{4dt}]$

Rate of disappearance of $O_2$ = $\frac{-\Delta [O_2]}{5dt}$

Rate of appearance of $NO_2$ = $\frac{+\Delta [NO_2]}{4dt}$

Rate of appearance of $H_2O$ = $\frac{+\Delta [H_2O]}{6dt}$