(rate constant K is equal to collision frequency A if E_a = 0)
In zero order reactions, increase in the concentration of reactant does not alter the rate.
So statement (i) is wrong.
\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
if E_a = 0 so, statement (ii) is correct, and statement (iii) is wrong
k = Ae⁰
k = A
In k = In A - \({ \left( \frac { { E }_{ a } }{ R } \right) }\) \(\left( \frac { 1 }{ T } \right) \)
This equation is of the form of a straight line y = mx+c
A plot of In k Vs \(\left( \frac { 1 }{ T } \right) \) gives a straight line with a negative slope
So statement (iv) and (v) are wrong.

Rate = k[A]ⁿ
Rate = \(\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\)
unit of rate = \(\frac{mol L^{-1}}{s}\)=mol L^-1/s^-1
unit of rate constant 
\(=\frac{ (mol{ L }^{ -1 }{ S }^{ -1 }) }{ ({ mol }{ L }^{ -1 })^n } \)
= mol^1-nL^n-1s^-1
in the case
rate = k [Acetone]^3/2
n = 3/2
= mol^1-(3/2)L^(3/2)-1s^-1
(mol^-(1/2) L^(1/2) s^-1).

 \(Rate=\frac { -1 }{ 2 } \frac { d[NH_3] }{ dt } \)
\(\frac { -d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { H}_{ 2 } \right] }{ dt } \)
\(=\frac { 1 }{ 2 } { k_1\left[ { NH}_{ 3 } \right] } = { k_2 \left[ { NH}_{ 3 } \right] }=\frac { 1 }{ 3 } { k_3\left[ { NH}_{ 3 } \right] } \)
[3/2] k₁ = 3k₂ = k₃
1.5k₁ = 3k₂ = k₃

\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
In k = In A - \({ \left( \frac { { E }_{ a } }{ R } \right) }\) \(\left( \frac { 1 }{ T } \right) \)
This equation is of the form of a straight line y = mx+c
A plot of In k Vs \(\left( \frac { 1 }{ T } \right) \) gives a straight line with a negative slope.

\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(k=\frac { 1 }{ t } ln \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(e^{kt}=\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] }\)
[A] = [A₀] e^kt
In this case
k = x min^-1 and [A₀]  = 0.01 M = 1 x 10^-2M
t = 1 hour = 60 min
[A] = (1 x 10^-2)e^-60x