First Order Reaction
Trending Questions
(Given that log2 =0.3010)
- 230.3 s
- 301 s
- 602 s
- 2000 s
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
A radioactive nucleus decays by two different processes. The half-life for the first process is and that for the second is . The effective half-life of the nucleus is close to:
A first order reaction takes 40 min for 30% decomposition. Calculate t12
- 0.69 / k
- 0.75 / k
- 0.10 / k
- 0.29 / k
Which of the following graphs is correct for a first order reaction ?
- 6.93×10−4 mol min−1
- 2.66 L min−1 at STP
- 1.34×10−2 mol min−1
- 6.93×10−2 mol min−1
- C6H6
- C6H5−C6H5
- C6H12
- C6H5−O−C6H5
Consider a first order gas phase decomposition reaction given below
A(g)→B(g)+C(g) The initial prssure of the system before decomposition of A was pi. After lapse of time 't' total pressure of the system increased by x units and became 'pt'. The rate constant k for the reaction is given as ...... .
(a)k=2.303tlogpipi−x(b)k=2.303tlogpi2pi−pt(c)k=2.303tlogpi2pi+pt(d)k=2.303tlogpipi−x
- 10 s
- 10000 s
- 1000 s
- 100 s
What is the rate law for a first-order reaction?
A first order reaction is 50% completed in 1.26×1014s. How much time would it take for 100% completion ?
(a)1.26×1015s(b)2.52×1014s(c)2.52×1028s(d)Infinite
If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately
- 40 min
- 50 min
- 60 min
- 45 min
- 1tlnP0Pt−P0
- 1tln3P0Pt−P0
- 1tln3P03Pt−P0
- 1tln2P03P0−Pt
Give the unit of Rate Constant of a Reaction.
The atomic mass of silver found in nature is . This silver is made of and Isotopes. Calculate the proportion of the mass of isotope which is in natural silver atoms.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t12=3.00h. What fraction of sample of sucrose remains after 8h?
The results given in the below table were obtained during kinetic studies of the following reaction: 2A + B → C + D
X and Y in the given table are respectively :
0.4, 0.4
0.3, 0.4
0.4, 0.3
0.3, 0.3
106 dynes cm-2
102 dynes cm-2
104 dynes cm-2
108 dynes cm-2
- 120 min
- 299 min
- 399 min
- 320 min
Column IColumn IIa)V0 is proportional top) Total concentrations of acid initiallypresent and concentration of acidformed at time t.b)Vt is proportional toq) Concentration of acid initiallypresent as the catalystc)(V∞−Vt) is proportional tor) Concentration of acid formed afterthe completion of reactiond)(V∞−V0)is proportional tos) Concentration of ester remainingat time t.
- (a-q, b-p, c-s, d-r)
- (a-p, b-p, c-s, d-r)
- (a-q, b-q, c-s, d-r)
- (a-q, b-q, c-r, d-s)
In the reaction, P + Q ⟶ R + S
the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is
- 0
- 1
2
- 3
Consider a certain reaction; A → products with k=2.0×10−2s−1. Calculate the concentration of A remaining after 100 s, if the initial concentration of A is 1.0 mol L−1.
- 100 minutes
- 200 minutes
- 300 minutes
- 30 minutes
Calculate the following:
Number of molecules of in 80.0 g of it.
The rate constant is
- 4.16×10−7s−1
- 2.5×10−5s−1
- 1.5×10−5s−1
- 1.5×10−3s−1
where C0 and C are the values of the reactant concentration at the start and after time t . What is the relationship between t3/4 and t1/2where t3/4 is the time required for to become 14C0?
- t3/4=t1/2[2n−1+1]
- t3/4=t1/2[2n−1−1]
- t3/4=t1/2[2n+1−1]
- t3/4=t1/2[2n+1+1]
- 10−2
- 10−4
- 10−5
- 10−6
A→B+C, rate constant is 0.001 M/s. If we start with 1 M of A, the concentration of A and B after 10 min are respectively:
- 0.5M, 0.5M
- 0.4M, 0.6M
- 0.6M, 0.4M
- None of these