According to Arrhenius equation rate constant K is equal to Ae^{\frac{-E_a}{RT}} which of the following options represents the graph of ln(K) vs 1/T?

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 

Answers (1)

As we learned in concept

Effect of Temperature and Activation Energy on rate constant -

E_{a}= 0                k\neq f(T)

E_{a}< 0                k decreases with the increase in T

 

 

E_{a}> 0                 k increases with increase in T

E_{a} = Activation Energy

T = Temperature

 k = Rate constant

- wherein

 

 

 

 

 

Effect of Temperature and Activation Energy on rate constant -

E_{a}= 0                k\neq f(T)

E_{a}< 0                k decreases with the increase in T

E_{a}> 0                 k increases with increase in T

E_{a} = Activation Energy

T = Temperature

 k = Rate constant

- wherein

 

Arrhenius equation is given by 

 

k=Ae^-{\frac{Ea}{Rt}}

\ln k= \ln A- \frac{Ea}{RT}

\ln k= \ln A-\frac{Ea}{R}(\frac{1}{T})

The graph of the above equaion can be drawn as follows:


Option 1)

Correct option

Option 2)

Incorrect option

Option 3)

Incorrect option

Option 4)

Incorrect option

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