22378.Half life period of a first order reaction is 1386 seconds. The specific rate constant of the reaction is
0.5 × 10–2 s–1
0.5 × 10–3 s–1
5.0 × 10–2 s–1
5.0 × 10–3 s–1
22379.Which one of the following is a second order reaction?
CH3COOCH3 + NaOH → CH3COONa + H2O
H2 + Cl2 $\xrightarrow{ \; \; \;sunlight \; \; \;}$ 2HCl
NH4NO3 → N2 + 3H2O
H2 + Br2 → 2BHr
22380.The time for half life period of a certain reaction A → Products is 1 h. When the initial concentration of the reactant 'A' is 2.0 mol L–1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L–1, if it is a zero order reaction?
0.25 h
1 h
4 h
0.5 h
22381.The reaction: A → B follows first order kinetics. The time taken for 0.8 mol of A to produce 0.6 mol of B is 1 hour. What is the time taken for conversion of 0.9 mol of A to produce 0.675 mol of B?
1 hour
0.5 hour
0.25 hour
2 hour
22382.In the reaction
BrO–
3 (aq) + 5Br– (aq) + 6H+ → 3Br2 (l) + 3H2O(l)
The rate of appearance of bromine (Br2) is related to the rate of disappearance of bromide ions as
$\dfrac{\text{d}[Br_2]}{\text{d}t}=–\dfrac{5}{3}\dfrac{\text{d}[Br^–]}{\text{d}t}$
$\dfrac{\text{d}[Br_2]}{\text{d}t}=\dfrac{5}{3}\dfrac{\text{d}[Br^–]}{\text{d}t}$
$\dfrac{\text{d}[Br_2]}{\text{d}t}=\dfrac{3}{5}\dfrac{\text{d}[Br^–]}{\text{d}t}$
$\dfrac{\text{d}[Br_2]}{\text{d}t}=–\dfrac{3}{5}\dfrac{\text{d}[Br^–]}{\text{d}t}$
22383.A reaction was found to be second order with respect to concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will
triple
increase by a factor of 4
double
remain unchanged
22384.The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2 = 0.301)
230.3 minutes
23.03 minutes
46.06 minutes
460.6 minutes
22385.The activation energy for a reaction at the temperature T K was found to be 2.303 RT J mol–1. The ratio of the rate constant to Arrhenius factor is ______.
10–1
10–2
2 × 10–3
2 × 10–2
22386.In a zero–order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become
64 times
128 times
256 times
512 times
22387.The bromination of acetone that occurs in acid solution is represented by the equation
Initial Rate, disappearance of Br2, ms–1
(a)5.7 × 10–5
(b)5.7 × 10–5
(c)1.2 × 10–4
(d)3.1 × 10–4
Based on these data, the rate of reaction is
CH3COCH3(aq) + Br2(aq) → CH3COCH2Br(aq) + H+(aq) + Br–(aq)
These kinetic data were obtained from given reaction concentrations.
Initial concentrations, (M)
| [CH3COCH3] | [Br2] | [H+] |
| 0.30 | 0.05 | 0.05 |
| 0.30 | 0.10 | 0.05 |
| 0.30 | 0.10 | 0.10 |
| 0.40 | 0.05 | 0.20 |
Initial Rate, disappearance of Br2, ms–1
(a)5.7 × 10–5
(b)5.7 × 10–5
(c)1.2 × 10–4
(d)3.1 × 10–4
Based on these data, the rate of reaction is
rate = k [CH3COCH3] [Br2]
rate = k [CH3COCH3] [Br2] [H+]2
rate = k [CH3COCH3] [Br2] [H+]
rate = k [CH3COCH3] [H+]
22388.Which of these does not influence the rate of reaction?
Nature of the reactants
Concentration of the reactants
Temperature of the reaction
Molecularity of the reaction
22389.The rate constants k1 and k2 for two different reactions are 1016 × e–2000/T and 1015 × e–1000/T, respectively. The temperature at which k1 = k2 is
$\dfrac{2000}{2.303}$ K
2000 K
$\dfrac{1000}{2.303}$ K
1000 K
22390.Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively
sec–1, M sec–1
sec–1, M
M sec–1, sec–1
M, sec–1
22391.If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately
45 minutes
60 minutes
40 minutes
50 minutes
22392.In a first order reaction A → B, if k is rate constant and initial concentration of the reactant A is 0.5 M, then the half–life is
$\dfrac{\text{log 2}}{\text{k}}$
$\dfrac{\text{log 2}}{\text{k}\sqrt{0.5}}$
$\dfrac{\text{ln 2}}{\text{k}}$
$\dfrac{\text{0.693}}{\text{0.5k}}$
22393.Consider the reaction:
N2(g) + 3H2(g) → 2NH3(g)
The equality relationship between $\dfrac{\text{d}[NH_3]}{\text{d}t}$ and $\dfrac{\text{d}[H_2]}{\text{d}t}$ is
+ $\dfrac{\text{d}[NH_3]}{\text{d}t}$ = – $\dfrac{2}{3}\dfrac{\text{d}[H_2]}{\text{d}t}$
+ $\dfrac{\text{d}[NH_3]}{\text{d}t}$ = – $\dfrac{3}{2}\dfrac{\text{d}[H_2]}{\text{d}t}$
+ $\dfrac{\text{d}[NH_3]}{\text{d}t}$ = – $\dfrac{\text{d}[H_2]}{\text{d}t}$
+ $\dfrac{\text{d}[NH_3]}{\text{d}t}$ = – $\dfrac{1}{2}\dfrac{\text{d}[H_2]}{\text{d}t}$
22394.For the reaction: 2A + B → 3C + D, which of the following does not express the reaction rate?
– $\dfrac{\text{d}[B]}{\text{d}t}$
$\dfrac{\text{d}[B]}{\text{d}t}$
–$\dfrac{\text{d}[A]}{\text{d}t}$
–$\dfrac{\text{d}[C]}{\text{3d}t}$
22395.The energies of activation for forward and reverse reactions for A2 + B2 ⇌ 2AB are 180 kJ mol–1 and 200 kJ mol–1 respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol–1. The enthalpy change of the reaction (A2 + B2 → 2AB) in the presence of catalyst will be (in kJ mol–1):
120
280
20
300
22396.Rate of a reaction can be expressed by Arrhenius equation as: k = Ae–E/RT In this equation, E represents
the energy below which colliding molecules will not react.
the total energy of the reacting molecules at a temperature, T.
the fraction of molecules with energy greater than the activation energy.
the energy above which all the colliding molecules will react.
22397.In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is
7.5 min
15 min
30 min
60 min