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For a reaction, consider the plot of In k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10-5 s-1 , then the rate constant at 500 K is :

  • Option 1)

    10^{-6}\: s^{-1}

  • Option 2)

    2 \times 10^{-4}\: s^{-1}

  • Option 3)

    10^{-4}\: s^{-1}

  • Option 4)

    4 \times10^{-4}\: s^{-1}

Answers (1)

best_answer

 

Chemical Kinetics -

The branch of chemistry which predicts the rate and mechanism of a process

-

 

 

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

 

 

 

Rate of reaction -

The  rate of the reaction during the courses of a reaction in any instant of time is the change in concentration of reacting species. Rate is a positive quantity.

Unit = mol \:L^{-} \:sec^{-}=\:[concentration][time]^{-}

- wherein

R\rightarrow P

t = t1

R_1\rightarrow P_1

t = t2

R_2\rightarrow P_2

\Delta t=t_2-t_1 \:\Delta R=[R_2]-[R_1]

\DeltaP=[P_2]-[P_1]

As we have learned

l_{n}K=l_{n}A-\frac{E_{a}}{R_{T}}=l_{n}A-\frac{4606}{T}

ln(\frac{k}{10^{-5}})=(\frac{Ea}{R})\times \frac{500-400}{500\times 400}

l_{n}\left ( \frac{K}{10^{-5}} \right )=4606\times \frac{1}{2000}=2.303

l_{n}\left ( \frac{K}{10^{-5}} \right )l_{n}10

K=10^{-4}

 

 


Option 1)

10^{-6}\: s^{-1}

Option 2)

2 \times 10^{-4}\: s^{-1}

Option 3)

10^{-4}\: s^{-1}

Option 4)

4 \times10^{-4}\: s^{-1}

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