Chemistry, asked by Gundasrivalli, 7 months ago

in 69.3 min, a first order reaction is 50% complete. How much reactants are left after 161 min? ​

Answers

Answered by madeducators6
5

Given:

in 69.3 min = t_{1}

reaction is 50% complete

order of reaction= first order

To Find:

reactants left after 161 min =?

Explanation:

Now, we know that 1st order reaction,

k = \dfrac{2.303}{t} log(\dfrac{C_{initial} }{C_{final} } ) and  k = \dfrac{0.693}{t_{1/2} }

here t = time taken

Now, for  t_{1}

since the reaction is 50% complete assume the initial concentration be 100% then the left concentration reacted will be 50%

k = \dfrac{0.693}{69.3} = 0.01 min-1     -----(1)

lly for 161 min

k = \dfrac{2.303}{161} log(\dfrac{100 }{100-c } ) ---------(2)

equating equation 1 and 2

0.01 =  \dfrac{2.303}{161} log(\dfrac{100 }{100-c } )

on solving

0.01 \times \dfrac{161}{2.303}  =  log(\dfrac{100 }{100-c } )

0.699  =  log(\dfrac{100 }{100-c } )

taking antilog to the base 10

5 = \dfrac{100}{100-c}

500 - 5c = 100

5c = 400

c = 80

thus 100-c = 20% reactants are left after 161 min

Answered by shujju
0

Answers

This is correct

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