First Order Reaction
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(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
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.
Give the unit of Rate Constant of a Reaction.
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