The rate of a first-order reaction is 0.04 mol l^-1 s^-1 at 10 seconds and 0.03 mol l^-1 s^-1 at 20 seconds after initiation of the reaction The half life period of the reaction is
amit singh , 8 Years ago
Grade 11
3 Answers
Vikas TU
Last Activity: 8 Years ago
U have to just put the values into the formulae for the first order eqn. that is:
K = (2.303/t2 – t1)log(Q1/Q2)
=(2.303/(10)llog(0.04/0.03)
After solving,
K = 0.02876 /min
thalf = 0.693/k = > 24 min approx.
Garima Shrivastava
Last Activity: 6 Years ago
Solution is all correct above just
K = (2.303/t2 – t1)log(Q1/Q2)
=(2.303/(10)llog(0.04/0.03)
After solving,
K= 2.6 / second*
And time = 24 seconds* approximately
Garima Shrivastava
Last Activity: 6 Years ago
K value went wrong. Uploading again. Sorry Solution is all correct above just
K = (2.303/t2 – t1)log(Q1/Q2)
=(2.303/(10)llog(0.04/0.03)
After solving,
K= 0.02 second*
And time = 24 seconds* approximately
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