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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Jun 2008 23:36:12 IST
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Q1) The activation energy of the reaction: 2HI (g) --------> H2 + I2 is 209.5 KJ/mol. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
Q2) The half life for radioactive decay of 14C C is 5730 years. A sample had 80% 14C . Estimate the age of the sample.
Q3)The rate constant for the decomposition of hydrocarbons is 2.418 * 10-5 at 546 K. If the energy of activation is 179.9 kJ/mol, what is the value of pre-exponential factor.
Q4)The decomposition of A into product has value of k as 4.5 * 103 /s at 10 degrees celsius and energy of activation 60 kJ/mol. At what temperature would k be 1.5 8 104 /s.
pls explain
thnx
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Jun 2008 23:56:32 IST
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q1---
fraction of molecules of reactants having energy equal to or greater than activation energy= e^(-E/RT)
put the values n get the answer
q3---
k=Ae^(-E/RT)
k=2.418 * 10^-5
E=179.9 kJ/mol
T=546 K
R=0.0821
now put these values and get the value of A
hope it helps...
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for the second one. the decay constant of the reaction is ln2/T where T is the half life of the reaction = 5730 years so letAo be the initial conc of carbon in the sample and At be the final so At = 0.8Ao and so usin that formula ln(Ao/0.8Ao)=(ln2/5730)t and so t comes out to be round abt 1910 years please correct me if i am wrong or nudge me if you have any doubts
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Jun 2008 00:04:48 IST
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fpr d last one
log (k2/k1)=Ea/(2.303R)*(T2 - T1/t2*t1)
here substitue k1 as d given value
and k2 we wanna find hope it helps
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Jun 2008 00:09:46 IST
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4 pretty same asw the third one
use the equation k=Ae^-E/RT
so for the process occurin at different temperature energy of activation will remain the same
k1=Ae^-E/RT1
k2=Ae^-E/RT2
so k1/k2=e^(E/R)(1/T2-1/T1)
so let k1=4.5*10^3
T1=283K
k2=1.5*10^4
T2=?
E=60kJ/mol=60000J/mol
solve and calculate the ans
it is a little lengthy calculation but if you have any doubts then please feel free to ask
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Jun 2008 00:19:31 IST
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ans to 4th one comes to 297.62K or 24.62C
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