Advertisements
Advertisements
Question
The rate constant for the first order decomposition of H2O2 is given by the following equation:
log k = 14.34 − 1.25 × 104 K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Advertisements
Solution
According to the Arrhenius equation,
k = `Ae^(-E_a//RT)`
or, loge k = `log_e A - E_a/(RT)`
or, log10 k = `log_10 A - E_a/(2.303 RT)` ...(i)
The given equation is:
log10 k = `14.34 - (1.25 xx 10^4 (K))/T` ...(ii)
Comparing equations (i) and (ii), we have,
`E_a/(2.303 RT) = (1.25 xx 10^4 (K))/T`
or, Ea = 2.303 × R × 1.25 × 104 (K)
= 2.303 × (8.314 × 10−3 kJ K−1 mol−1) × 1.25 × 104 (K)
= 239.34 kJ mol−1
When t1/2 = 256 min, i.e., 256 × 60 s, we have
k = `0.693/t_(1//2)`
= `0.693/(256 xx 60)`
= 4.51 × 10−5 s−1
Substituting this value in equation (ii), we have,
log10 (4.51 × 10−5) = `14.34 - (1.25 xx 10^4 (K))/T`
⇒ −4.346 = `14.34 - (1.25 xx 10^4 (K))/T`
⇒ `(1.25 xx 10^4 (K))/T` = 14.34 + 4.346
⇒ `(1.25 xx 10^4 (K))/T` = 18.686
T = `(1.25 xx 10^4 (K))/18.686`
= 668.95 K
RELATED QUESTIONS
The integrated rate equation for first order reaction is A → products
Derive the relation between half life and rate constant for a first order reaction
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
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
The experimental data for decomposition of N2O5
\[\ce{2N2O5 -> 4NO2 + O2}\]
in gas phase at 318 K are given below:
| t/s | 0 | 400 | 800 | 1200 | 1600 | 2000 | 2400 | 2800 | 3200 |
| 102 × [N2O5]/mol L−1 | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35 |
- Plot [N2O5] against t.
- Find the half-life period for the reaction.
- Draw a graph between log [N2O5] and t.
- What is the rate law?
- Calculate the rate constant.
- Calculate the half-life period from k and compare it with (ii).
During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically?
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))`
In a first-order reaction A → product, 80% of the given sample of compound decomposes in 40 min. What is the half-life period of the reaction?
The half life period of a radioactive element is 140 days. After 560 days, 1 g of element will be reduced to
Show that for a first order reaction half life is independent of initial concentration.
The rate constant for a first order reaction is 1.54 × 10−3 s−1. Calculate its half life time.
Calculate half-life period of life order reaction whose rate constant is 200 sec–1
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)
Observe the graph shown in figure and answer the following questions:
Write the relationship between k and t1/2 (half-life period)
A sample of U238 (half-life = 4.5 × 109 years) ore is found to contain 23.8 g of U238 and 20.6 g of Pb206. The age of the ore is ______ × 109 years.
The amount of C-14 isotope in a piece of wood is found to be 1/16th of its amount present in a fresh piece of wood. The age of wood, half-life period of C-14 is 5770 years, is ______ years.
The half-lives of a first-order reaction are 1.19s at 313 K and 15.45s at 293 K. Calculate the energy of activation.
Show that the half-life of zero order reaction is `t_(1/2) = ([A]_0)/(2k)`.
