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Question: The decomposition of \(N{{H}_{3}}\) on a Pt surface is a zero-order reaction. If the value of rate c...

The decomposition of NH3N{{H}_{3}} on a Pt surface is a zero-order reaction. If the value of rate constant is 2×10−4mole litre−1sec⁡−12\times {{10}^{-4}}mole\text{ litr}{{\text{e}}^{-1}}{{\sec }^{-1}}. The rate of appearance of N2{{N}_{2}} and H2{{H}_{2}} are respectively.
(A) N=1×10−4mol l−1sec⁡−1,H2=3×10−4mol l−1sec⁡−1N=1\times {{10}^{-4}}mol\text{ }{{\text{l}}^{-1}}{{\sec }^{-1}},{{H}_{2}}=3\times {{10}^{-4}}mol\text{ }{{\text{l}}^{-1}}{{\sec }^{-1}}
(B) N=3×104mol l−1sec⁡−1,H2=1×10−4mol l−1sec⁡−1N=3\times {{10}^{4}}mol\text{ }{{\text{l}}^{-1}}{{\sec }^{-1}},{{H}_{2}}=1\times {{10}^{-4}}mol\text{ }{{\text{l}}^{-1}}{{\sec }^{-1}}
(C) N=2×10−4mol l−1sec⁡−1,H2=6×10−4mol l−1sec⁡−1N=2\times {{10}^{-4}}mol\text{ }{{\text{l}}^{-1}}{{\sec }^{-1}},{{H}_{2}}=6\times {{10}^{-4}}mol\text{ }{{\text{l}}^{-1}}{{\sec }^{-1}}
(D) N=3×10−4mol l−1sec⁡−1,H2=3×10−4mol l−1sec⁡−1N=3\times {{10}^{-4}}mol\text{ }{{\text{l}}^{-1}}{{\sec }^{-1}},{{H}_{2}}=3\times {{10}^{-4}}mol\text{ }{{\text{l}}^{-1}}{{\sec }^{-1}}

Explanation

Solution

We come to find that in some reactions the rate is found to be independent of the reactant concentration. The rates of zero-order reactions do not change with increasing nor decreasing reactant’s concentrations. Thus, the rate of the reaction is equal to the rate constant of the reaction which differs from first-order reactions and second-order reactions.

Complete answer:
Let’s introduce the term Chemical Kinetics.
Chemical kinetics is a topic in Physical Chemistry that deals with helping students know the different aspects of a chemical reaction. In addition, the term ‘kinetics’ deals with the rate of change of quantity.
You would be wondering what are the rate of formations and disappearances? Let’s check it out.
In a chemical reaction, as the reaction is found to proceed the amount of the reactants decreases and the amount of products increases. One must understand that the rate of overall reaction depends on the rate at which the rate at which the products are formed, or the reactants are consumed.
The rate of formation of products and rate of disappearance of reactants can be calculated from the slope of curves for products and reactants, if a graph is plotted between the concentration of reactants and products and time.
The total rate of reaction may be or may also not be equal to the rate of formations and disappearances.
For the given question let’s write the equation.
2NH3→N2+3H22N{{H}_{3}}\to {{N}_{2}}+3{{H}_{2}}
So, the rate of reaction for decomposition of ammonia can be given as follows:
dKdt=−12d(NH3)dt=d(N2)dt=13d(H2)dt\dfrac{dK}{dt}=-\dfrac{1}{2}\dfrac{d(N{{H}_{3}})}{dt}=\dfrac{d({{N}_{2}})}{dt}=\dfrac{1}{3}\dfrac{d({{H}_{2}})}{dt}
Where K is the rate constant.
So, the rate of appearance of N2=dKdt=d(N2)dt=2×10−4mole litre−1sec⁡−1{{N}_{2}}=\dfrac{dK}{dt}=\dfrac{d({{N}_{2}})}{dt}=2\times {{10}^{-4}}mole\text{ litr}{{\text{e}}^{-1}}{{\sec }^{-1}}
And the rate of appearance of H2=d(H2)dt=3×dKdt=3×2×10−4mole litre−1sec⁡−1=6×10−4mole litre−1sec⁡−1{{H}_{2}}=\dfrac{d({{H}_{2}})}{dt}=3\times \dfrac{dK}{dt}=3\times 2\times {{10}^{-4}}mole\text{ litr}{{\text{e}}^{-1}}{{\sec }^{-1}}=6\times {{10}^{-4}}mole\text{ litr}{{\text{e}}^{-1}}{{\sec }^{-1}}

Therefore, the correct option is (C) .

Note:
We should keep in mind that the coefficients if the product and the reactants should not be left as that would result in different answers. Thus, special care needs to be taken while writing the rate of reaction for decomposition of ammonia or any other chemical equation.