Advertisements
Advertisements
प्रश्न
The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 1010 s−1. Calculate k at 318 K and Ea.
Advertisements
उत्तर
Let [R]0 = 100
t = `2.303/k_1 log 100/90 = 2.303/k_2 log 100/75`
⇒ `2.303/k_1 log 10/9 = 2.303/k_2 log 4/3`
⇒ `1/k_1 log 1.111 = 1/k_2 log 1.333`
⇒ `k_2/k_1 = (log 1.333)/(log 1.111)`
⇒ `k_2/k_1 = 0.1249/0.0457`
⇒ `k_2/k_1` = 2.733
⇒ `log k_2/k_1` = log 2.733 = 0.4367
`log k_2/k_1 = E_a/(2.303 R) ((T_2 - T_1)/(T_1T_2))`
⇒ 0.4367 = `E_a/(2.303 xx 8.314) ((308 - 298)/(298 xx 308))`
⇒ Ea = `(2.303 xx 8.314 xx 308 xx 298)/(10) xx 0.4367`
= `(19.147 xx 308 xx 298)/(10) xx 0.4367`
= 76743.13 J mol−1
⇒ Ea = `76743.13/1000`
= 76.74 kJ mol−1
log k = `log A - E_a/(2.303 RT)`
⇒ log k = `log (4 xx 10^10) - (76.74 xx 1000)/(2.303 xx 8.314 xx 318)`
⇒ log k = `10.6021 - 76740/6088.79`
⇒ log k = 10.6021 − 12.6034
⇒ log k = −2.0013
⇒ k = Antilog(−2.0013)
= Antilog(3.9987)
⇒ k = 9.97 × 10−3 s−1
Notes
The answer in the textbook is incorrect.
संबंधित प्रश्न
In a pseudo first order reaction in water, the following results were obtained:
| t/s | 0 | 30 | 60 | 90 |
| [A]/mol L−1 | 0.55 | 0.31 | 0.17 | 0.085 |
Calculate the average rate of reaction between the time interval 30 to 60 seconds.
The rate constant for a first order reaction is 60 s−1. How much time will it take to reduce the initial concentration of the reactant to its `1/16`th value?
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
| t (sec) | P(mm of Hg) |
| 0 | 35.0 |
| 360 | 54.0 |
| 720 | 63.0 |
Calculate the rate constant.
Following data are obtained for reaction :
N2O5 → 2NO2 + 1/2O2
| t/s | 0 | 300 | 600 |
| [N2O5]/mol L–1 | 1.6 × 10-2 | 0.8 × 10–2 | 0.4 × 10–2 |
1) Show that it follows first order reaction.
2) Calculate the half-life.
(Given log 2 = 0.3010, log 4 = 0.6021)
Straight line graph for first order reaction is obtained between ____________.
State a condition under which a bimolecular reaction is kinetically first order reaction.
In the presence of acid, the initial concentration of cane sugar was reduced from 0.2 M to 0.1 Min 5 hours and to 0.05 Min 10 hours. The reaction must be of?
First order reaction is 50% complete in 1.26 × 1014s. How much time could it take for 100% completion?
In the first order reaction, half of the reaction is complete in 100 seconds. The time for 99% of the reaction to occurs will be
The reaction X → product
Follow first order of kinetics. In 40 minutes the concentration of 'X' changes from 0.1 m to 0.025. M. The rate of reaction when concentration of X is 0.01 m is.
Observe the graph shown in figure and answer the following questions:
- What is the order of the reaction?
- What is the slope of the curve?
- Write the relationship between k and t1/2 (half life period).
The reaction \[\ce{SO2Cl2(g) -> SO2(g) + Cl2(g)}\] is a first-order gas reaction with k = 2.2 × 10−5 sec−1 at 320°C. The percentage of SO2Cl2 is decomposed on heating this gas for 90 min, is ______%.
The slope in the plot of `log ["R"]_0/(["R"])` Vs. time for a first-order reaction is ______.
How will you represent first order reactions graphically?
Slove: \[\ce{2NOBr -> 2NO_{2(g)} + Br_{2(g)}}\]
For the above reaction, the rate law is rate = k[NOBr]2. If the rate of reaction is 6.5 × 10−6 mol L−1 s−1 at 2 × 10−3 mol L−1 concentration of NOBr, calculate the rate constant k for the reaction.
The rate constant for the reaction:
\[\ce{2N2O_{(s)} ->2N2O4_{(g)}}\] is 4.98 × 10-4 s-1.
The order of the reaction is ______.
Write the equation for integrated rate law for a first order reaction.
Show that `t_(1/2)= 0.693/k` for first reaction.
