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NEET Class neet Answered

how does the ratio of the natural log of K 37 degree Celsius to K 27 degree Celsius in this problem equal 1 ? 
question image
Asked by astutijoshi | 05 Jul, 2019, 06:17: PM
answered-by-expert Expert Answer
Option (1) is correct.
 
Given:
 
T1 = 27 °C
 
T2 =37 °C
 
Ea = 83 kJ
 
According to Arrhenius equation,
 
ln space straight k space equals space ln space straight A space minus Ea over RT

For space two space different space temperatures comma space straight T subscript 1 space and space straight T subscript 2

ln open parentheses straight k subscript 1 over straight k subscript 2 close parentheses space equals space Ea over straight R open square brackets 1 over straight T subscript 2 minus 1 over straight T subscript 1 close square brackets


ln open parentheses straight k subscript 1 over straight k subscript 2 close parentheses space equals space space Ea over straight R open square brackets fraction numerator straight T subscript 1 minus straight T subscript 2 over denominator straight T subscript 1 straight T subscript 2 end fraction close square brackets

ln open parentheses straight k subscript 37 over straight k subscript 27 close parentheses space equals fraction numerator 83 cross times 10 cubed over denominator 8.314 end fraction open square brackets fraction numerator negative 10 over denominator 93000 end fraction close square brackets

space space space space space space space space space space space space space space equals fraction numerator 83 over denominator 8.314 end fraction cross times 0.1075

space space space space space space space space space space space space space space equals fraction numerator 8.9247 over denominator 8.314 end fraction

space space space space space space space space space space space space space space space equals space 1.07
ln open parentheses straight k subscript 37 over straight k subscript 27 close parentheses space almost equal to space 1
 
The ratio of ln(k37 / k27 ) is 1.
 
Answered by Varsha | 09 Jul, 2019, 02:20: PM
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