Question

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

Answer 1

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

Answer 2

Answers

This is correct

Explanation: