Sunday 9 June 2013

A first order reaction has a rate constant 1.15 x 10−3 s−1. How long will 5 g of this reactant take to reduce to 3 g?


Answer
Given that
Initial quantity, [R]o= 5 g
Final quantity, [R] = 3 g
Rate constant, k = 1.15 x 10−3 s−1
Formula of 1st order reaction,

Plug the values we get

Value of log 5/3 = log 5 – log 3 = 0.2219
We get

After calculation we get
444 sec
Hence, it will take 444 sec to reduce 5 g of this reactant to 3 g

Do you know any other method of solving this problem ?
A first order reaction has a rate constant 1.15 x 10−3 s−1. How long will 5 g of this reactant take to reduce to 3 g?
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8 comments:

  1. Einstein Foundation15 November 2017 at 02:35

    Cannot we do it like log5/3 = log of 1.66 or simply log2 i.e., 0.204?

    ReplyDelete
    Replies
    1. Unknown9 April 2019 at 09:13

      Huge error 1.66<2.0 ur approx. Is hell large

      Delete
  2. Unique Social Media Boost3 April 2018 at 03:48

    We can but log of 5 - log of 3 is more easy because... We all know the values of log 5 and 3..... To find Log of 1.66 we wud need the log table

    ReplyDelete
  3. Unknown20 September 2018 at 02:43

    We cannot take log2 because both log5 and log3 has not the exact value as a numerical three and five......Otherwise why we are taking the heck of this freaking 'logs'😺

    ReplyDelete
  4. Unknown9 April 2019 at 09:10

    But you are not taking concentrations you are taking this quantity in grams not in mol per liter per time

    ReplyDelete
  5. Unknown7 July 2019 at 09:41

    why we not use t=0.693/k*log[R]o/[R]

    ReplyDelete

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