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# For the reaction, $4N{H_3} + 5{O_2} \to 4NO + 6{H_2}O$, if the rate of disappearance of $N{H_3}$ is $3.6 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}}$, what is the rate of formation of ${H_2}O$?(A) $5.4 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}}$(B) $3.6 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}}$(C) $4 \times {10^{ - 4}}mol{L^{ - 1}}{s^{ - 1}}$(D) $0.6 \times {10^{ - 4}}mol{L^{ - 1}}{s^{ - 1}}$

Last updated date: 13th Jun 2024
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Hint: Try to recall that rate of reaction is defined as the change in any one of the reactants or products per unit time. Now, by using this you can easily find the correct option from the given one.

Complete step by step solution:
It is known to you that the rate of a reaction can be expressed in terms of any reactant or product.
As concentration of reactant decreases, a negative sign is used to express the rate of reaction in terms of reactants.
As concentration of products increases, a positive sign is used to express the rate of reaction in terms of products.
Also, to get a unique value of the reaction rate (independent of the concentration terms chosen), we divide the rate of reaction defined with any of the reactants or products by the stoichiometric coefficient of that reactant or product involved in the reaction.
For the reaction, $4N{H_3} + 5{O_2} \to 4NO + 6{H_2}O$
Rate of reaction $= - \dfrac{1}{4}\dfrac{{\Delta \left[ {N{H_3}} \right]}}{{\Delta t}} = - \dfrac{1}{5}\dfrac{{\Delta \left[ {{O_2}} \right]}}{{\Delta t}} = \dfrac{1}{4}\dfrac{{\Delta \left[ {NO} \right]}}{{\Delta t}} = \dfrac{1}{6}\dfrac{{\Delta \left[ {{H_2}O} \right]}}{{\Delta t}}$
Given, rate of disappearance of $N{H_3}$=$- \dfrac{{\Delta \left[ {N{H_3}} \right]}}{{\Delta t}}$=$3.6 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}}$.
Rate of formation of ${H_2}O$=$\dfrac{{\Delta \left[ {{H_2}O} \right]}}{{\Delta t}}$.
From rate equation,
$\Rightarrow \dfrac{1}{4}\dfrac{{\Delta \left[ {N{H_3}} \right]}}{{\Delta t}} = \dfrac{1}{6}\dfrac{{\Delta \left[ {{H_2}O} \right]}}{{\Delta t}}$
$\therefore \dfrac{{\Delta \left[ {{H_2}O} \right]}}{{\Delta t}} = \dfrac{6}{4} \times 3.6 \times {10^{ - 3}} = 5.4 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}}$.

Hence, from above we can conclude that option A is the correct option to the given question.

Note:
- It should be remembered to you that in aqueous solutions, the rate of a reaction is not expressed in terms of change of concentration of water because the change is very small and negligible.
- Also, you should remember that the plot of concentration of reactant vs time, the tangent at any instant of time has a negative solution.