For a reaction, the energy of activation is zero. The value of the rate constant at 300 K is if k = 1.6 x $10^{ 6 }s^{ -1 }$ at 280 K :
A.) 1.60 x $10^{ 6 }s^{ -1 }$
B.) 1.56 x $10^{ 6 }s^{ -1 }$
C.) 1.82 x $10^{ 6 }s^{ -1 }$
D.) None of these

Hint: To find the answer you have to use the Arrhenius equation and from there you can easily calculate the value of the rate constant at 300 K as you already have the value at 280 K.

Complete step by step answer:

We know that the Arrhenius equation is represented as,
$k\quad =\quad Ae^{ -Ea/RT }$
When we have two rate constants at two different temperatures. We can also write this as,
$log\dfrac { { k }_{ 2 } }{ { k }_{ 1 } } \quad =\quad \dfrac { { E }_{ a } }{ 2.303 } \left( \dfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right)$
Here,
${ k }_{ 2 }$ = we have to calculate
${ k }_{ 1 }$ = 1.6 x $10^{ 6 }s^{ -1 }$
${ T }_{ 2 }$ = 300 K
${ T }_{ 1 }$ = 280 K
${ E }_{ a }$ = 0
Now, just insert the values in the above equation -
$\Rightarrow log\dfrac { { k }_{ 2 } }{ { k }_{ 1 } } \quad =\quad \dfrac { 0 }{ 2.303 } \left( \dfrac { 300-280 }{ 280\quad \times \quad 300 } \right) $
$\Rightarrow log\dfrac { { k }_{ 2 } }{ 1.6\quad \times \quad 10^{ 6 } } \quad =\quad 0$
$\Rightarrow \dfrac { { k }_{ 2 } }{ 1.6\quad \times \quad 10^{ 6 } } \quad =\quad antilog(0)$
$\Rightarrow \dfrac { { k }_{ 2 } }{ 1.6\quad \times \quad 10^{ 6 } } \quad =\quad 1$
$\Rightarrow { k }_{ 2 }$ = 1.6 x $10^{ 6 }s^{ -1 }$ = ${ k }_{ 1 }$
Here we can see that this value of the rate constant at 280 K is exactly the same as the value of the rate constant at 300 K. So, we can say that there is no effect.


Therefore, we can conclude that the correct answer to this question is option A.

Note: We should know the relation of activation energy with temperature - As temperature increases, molecules gain energy and move faster and faster. Therefore, the greater the temperature, the higher the probability that molecules will be moving with the necessary activation energy for a reaction to occur upon collision.