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प्रश्न
A first order reaction takes 40 min for 30% decomposition. Calculate `"t"_(1/2)`.
उत्तर
Remaining quantity = (a − x)
= a − 0.30a
= 0.70a
k = `2.303/"t" log "a"/("a" - "x")`
= `2.303/40 log "a"/(0.70"a")`
= `2.303/40 log 1.428`
k = 8.90 × 10−3 min−1
`"t"_(1/2)` = `0.693/(8.90 xx 10^-3 "min"^-1)`
= 77.7 min
संबंधित प्रश्न
Derive the relation between half life and rate constant for a first order reaction
A first order reaction takes 40 minutes for 30% decomposition. Calculate t1/2 for this reaction. (Given log 1.428 = 0.1548)
A first order reaction takes 23.1 minutes for 50% completion. Calculate the time required for 75% completion of this reaction.
(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
The unit of rate constant for zero order reaction is _______.
(A) t–1
(B) mol dm–3 t–1
(C) mol–1 dm3 t–1
(D) mol–2 dm6 t–1
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 zero order reaction A → product is given by
(a) `([A]_0)/k`
(b) `0.693/k`
(c) `[A]_0/(2k)`
(d) `(2[A]_0)/k`
The half life period of a first order reaction is 6. 0 h . Calculate the rate constant
Calculate half-life period of life order reaction whose rate constant is 200 sec–1
Assertion (A): The half-life of a reaction is the time in which the concentration of the reactant is reduced to one-half of its initial concentration.
Reason (R): In first-order kinetics, when the concentration of reactant is doubled, its half-life is doubled.
Show that the half-life of zero order reaction is `t_(1/2) = ([A]_0)/(2k)`.
