What percentage of the initial concentration of reactant reacts in 2.0 hours for a reaction whose rate constant is 4.25 x 10-5 sec ?
Answers
Answered by
2
Answer:
ANSWER
Given, K=6.8×10
−4
s
−1
; [A
o
]=0.04 M;
t=20 min=20×60=1200 sec
K=
t
2.303
log
[A
t
]
1
[A]
o
⟹6.8×10
−4
=
1200
2.303
log
10
(
[A
t
]
0.04
)
⟹Antiog(0.3545)=
[A
t
]
0.04
⟹2.261=
[A
t
]
0.04
⟹[A
t
]=0.0177 M
Similar questions