Question

Ammonium ion reacts with Nitrite ion according to the equation,
${{(N{{H}_{4}})}^{+}}_{(aq)}+{{(N{{O}_{2}})}^{-}}_{(aq)}\to {{N}_{2(g)}}+2{{H}_{2}}{{O}_{(l)}}$
The following initial rates of reaction have been measured for the given reactant concentrations.

Expt. No	${{(N{{H}_{4}})}^{+}}_{(aq)}M$	${{(N{{O}_{2}})}^{-}}_{(aq)}M$	Rate(M/hr)
1	0.010	0.020	0.020
2	0.015	0.020	0.030
3	0.010	0.010	0.005

Which of the following is the rate law for this reaction?
(A) $Rate=k\left[ N{{H}_{4}}^{+} \right]{{\left[ N{{O}_{2}} \right]}^{4}}$
(B) $Rate=k\left[ N{{H}_{4}}^{+} \right]\left[ N{{O}_{2}} \right]$
(C) $Rate=k\left[ N{{H}_{4}}^{+} \right]{{\left[ N{{O}_{2}} \right]}^{2}}$
(D) $Rate=k{{\left[ N{{H}_{4}}^{+} \right]}^{2}}\left[ N{{O}_{2}} \right]$
(E) $Rate=k{{\left[ N{{H}_{4}}^{+} \right]}^{\dfrac{1}{2}}}{{\left[ N{{O}_{2}} \right]}^{\dfrac{1}{4}}}$

Answer 1

Create three equations of rate law of the reaction for three conditions using concentrations given in the question, then compare those equations and performing mathematical steps will give the answer.

Complete step by step answer:
We can write a general formula for the rate of this reaction. Then we will make three equations for three different conditions given in the question. Then comparing those equations will lead us to the answer.
General formula for the rate of this reaction can be expressed as under.
$Rate=k{{\left[ N{{H}_{4}}^{+} \right]}^{x}}{{\left[ N{{O}_{2}} \right]}^{y}}$……………..(1)
where k is the rate constant.

We are not given the value of rate constant, so we will have to compare the three equations of three different conditions to derive the rate law equation.
We are given data of rate of reaction at different concentrations.
Now equation (1) for the experiment (1) will be
$0.020=k{{\left[ 0.010 \right]}^{x}}{{\left[ 0.020 \right]}^{y}}$………(2)
And equation (1) for experiment (2) will be
$0.030=k{{\left[ 0.015 \right]}^{x}}{{\left[ 0.020 \right]}^{y}}$………(3)
Dividing the equation (2) and (3) , we get
$\dfrac{0.020}{0.030}={{\left( \dfrac{0.010}{0.015} \right)}^{x}}{{\left( \dfrac{0.020}{0.020} \right)}^{y}}$
So,$\dfrac{2}{3}={{\left( \dfrac{2}{3} \right)}^{x}}$
So, x=1.

Now, equation (1) for experiment (3) will be
$0.005=k{{\left[ 0.010 \right]}^{x}}{{\left[ 0.010 \right]}^{y}}$……………..(4)
Dividing equation (2) and (4), we will get
$\dfrac{0.020}{0.005}={{\left( \dfrac{0.010}{0.010} \right)}^{x}}{{\left( \dfrac{0.020}{0.010} \right)}^{y}}$
$4={{\left( 2 \right)}^{y}}$
Hence, y=2

Now we will substitute the values of x and y obtained into the generalized rate law equation for this reaction to get the final equation for this reaction.
Substituting the values of x and y into equation (1), we get
$Rate=k{{\left[ N{{H}_{4}}^{+} \right]}^{1}}{{\left[ N{{O}_{2}} \right]}^{2}}$
So, the correct answer is “Option C”.

- There is an alternative way to solve this question.
- Compare experiment number (3) and (1), we can see that ammonium ion has the same concentrations and as nitrite ions’ concentration gets two times higher the rate becomes four times higher. This is only possible if this reaction is bimolecular with respect to nitrite ions.
- Comparing experiment (1) and (2), we can see that as the concentration of ammonium ions increases by one and a half time , the rate also increases by one and a half time. So, this is only possible if this reaction is unimolecular with respect to ammonium ions.

Note: - Do not put random values into random equations given in the options to get the question solved as it is very time consuming. There are two methods available. You can use any of them. Calculate mathematical steps in a step-by-step manner as it will make it easier for you to figure out mistakes if any.