
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
Answer
511.7k+ views
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.
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

