CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
232
You visited us 232 times! Enjoying our articles? Unlock Full Access!
Question

In a 1st order reac. A-> products the conc. Of reactant decreases to 6.25% of its initial value in 80 min. What is 1) the k 2)R,100minafter the start if the initial conc. Is 0.2 mole/litre

Open in App
Solution

a)
Let the initial concentration be R subscript 0
For first order reaction
k equals fraction numerator 2.303 over denominator t end fraction log R subscript 0 over R
Here,
R = 6.25R subscript 0
k equals fraction numerator 2.303 over denominator 80 end fraction log fraction numerator 100 R subscript 0 over denominator 625 R subscript 0 end fraction
k=0.0288 log100 over 625
k=0.03466 m i n to the power of negative 1 end exponent
b)
k equals fraction numerator 2.303 over denominator t end fraction log R subscript 0 over R
0.03466 = fraction numerator 2.303 over denominator 100 end fraction log fraction numerator 0.2 over denominator R end fractionrightwards double arrow 0.03466 cross times fraction numerator 100 over denominator 2.303 end fraction equals log fraction numerator 0.2 over denominator R end fraction
1.50499equals log fraction numerator 0.2 over denominator R end fraction
e to the power of 1.50499 end exponent equals fraction numerator 0.2 over denominator R end fraction
4.09098equals fraction numerator 0.2 over denominator R end fraction
R = 0.04888fraction numerator m o l e over denominator l i t e r end fraction

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Rate of Reaction
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon