Advertisements
Advertisements
प्रश्न
A first-order reaction takes 69.3 min for 50% completion. What is the time needed for 80% of the reaction to get completed? (Given: log 5 = 0.6990, log 8 = 0.9030, log 2 = 0.3010)
Advertisements
उत्तर
Half-life `t_(1/2) = 0.693/k`
k = `0.693/69.3 = 1/100` = 0.01 min–1
For first-order reaction
k = `2.303/t log ([Ro])/([R])`
t = `2.303/0.01 log 100/20`
t = 230.3 log 5 (log 5 = 0.6990)
t = 160.9 min
APPEARS IN
संबंधित प्रश्न
The integrated rate equation for first order reaction is A → products
The rate constant for a first order reaction is 100 s–1. The time required for completion of 50% of reaction is _______.
(A) 0.0693 milliseconds
(B) 0.693 milliseconds
(C) 6.93 milliseconds
(D) 69.3 milliseconds
A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
(log 2 = 0.3010)
Which among the following reactions is an example of a zero order reaction?
a) `H_(2(g)) + I_(2(g)) -> 2HI_(g)`
b) `2H_2O_(2(l)) -> 2H_2O_(l) + O_(2(g))`
c) `C_12H_22O_(11(aq)) + H_2O_(l) -> C_6H_12O_(6(aq)) + C_6H_12O_(6(aq))`
d) `2NH_(3g)` `N(2g) + 3H_(2(g))`
The half life period of a first order reaction is 6. 0 h . Calculate the rate constant
Define half life of a reaction.
For the first order reaction, half-life is equal to ____________.
Obtain a relation, `k_2/k_1 = ((t_(1/2))_2)/((t_(1/2))_1)`, where k1 and k2 are rate constants while (t1/2)1 and (t1/2)2 are half-life periods of the first order reaction at temperatures T1 and T2 respectively. Write the relation for activation energy.
Show that the half-life of zero order reaction is `t_(1/2) = ([A]_0)/(2k)`.
A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
