The rate of a reaction is expressed in different ways as follows :
The reaction is :
B + 3D = 2C + 4A
The rate constant for the forward reaction A(g) 2B(g) is 1.5 × 10-3 s-1 at 100 K. If 10-5 moles of A and 100 moles of B are present in a 10 litre vessel at equilibrium then rate constant for the backward reaction at this temperature is
Kf = = 1.5 × 10–3 s–1
At eq.m, Rf = Rb
Kf [A] = Kb[B]2
On solving : Kb = 1.5 × 10-11
Reaction A + B → C + D follow's following rate law : rate = k[A]+1/2[B]1/2. Starting with initial conc. of 1 M of A and B each, what is the time taken for concentration of A of become 0.25 M.
Given : k = 2.303 × 10-3 sec-1.
Overall order of reaction is = 1/2+1/2=1
∴ It is a 1st order reaction.
For concentration of A to become 0.5M:-t1/2=0.693/k
It will require 2 half times for the concentration of A to become .0.25M.
∴ Time taken= 2×t1/2 sec
Consider the following first order competing reactions :
X A+ B and y C+D
if 50% of the reaction of X was completed when 96% of the reaction of Y was completed, the ratio of their rate constants is
Units of rate constant for first and zero order reactions in terms of molarity (M) are respectively.
The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is raised by 50°C, the rate of the reaction increases by about
For every 10°C rise of temperature, the rate is doubled. Thus, the temperature coefficient of the reaction = 2
When temperature is increased by 50°, rate becomes
=2(50/10) = 25 times = 32 times
For a reaction pA + qB → products, the rate law expression is r = k[A]l[B]m, then :
pA + qB → Products r = k[A]l[B]m
(p + q) may or may not be equal to (l + m) (order may or may not be equal to stoichiometric coefficient for a complex reaction.
In the reaction : A + 2B → 3C + D, which of the following expression does not describe changes in the conc. of various species as a function of time :
A first order reaction is 87.5% complete in an hour. The rate constant of the reaction is
Half-life period of a second order reaction is
A first order reaction is 50% completed in 20 minutes at 27°C and in 5 min at 47°C. The energy of activation of the reaction is
t1/2 = 20 min at 300 K
t’1/2 = 5 min at 320 K
On solving, Ea = 55303.12 J
= 55.3 KJ
For the first order reaction A → B + C, carried out at 27°C if 3.8 × 10-16% of the reactant molecules exists in the activated state, the Ea (activation energy) of the reaction is
A → B + C
Ea = 100 KJ/mol
The rate constant, the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.0 × 10-4 s-1, 104.4 kJ mol-1 and 6.0 × 10-4s-1respectively. The value of the rate constant at T → ∞ is
At temperature = ∞
Rate constant = Arrhenious constant.
The following mechanism has been proposed for the exothermic catalyzed complex reaction.
If k1 is much smaller than k2. The most suitable qualitative plot of potential energy (P.E.) versus reaction coordinate for the above reaction.
K1 <<< K2 = most Imp. peack will be higher
The activation energy of a reaction at a given temperature is found to be 2.303 RT J mol–1. The ratio of rate constant to the Arrhenius factor is
Consider + heat, If activation energy for forward reaction is 100 kJ/mole then activation energy for backward reaction and heat of reaction is :
Eaf = 100 KJ/mol
Eab = ? heat of reaction = ?
Only one option because
Eb = Ea + ΔH.O. reaction
In a reaction, the thershold energy is equal to :
In a reaction, the threshold energy is equal to : Activation energy + Normal energy of reactants.
The first order rate constant k is related to temperature as log k = 15.0 - (106/T). Which of the following pair of value is correct ?
log A = 15
⇒ A = 1015
Ea = 1.9 × 104 KJ
When a graph between log K and 1/T is drawn a straight line is obtained. The temperature at which line cuts y-axis and x-axis.
When line cuts y axis
When it cut X -axis
log K = 0
The rate constant, the activation energy and the frequency factor of a chemical reaction at 25°C are 3.0 × 10-2 s-1, 104.4 KJ mol-1 and 6.0 × 1014 s-1 respectively. The value of the rate constant as T → ∞ is :
K = 3 × 10–2 s–1
Ea = 104.4 KJ/mol
A = 6 × 1014
Value of rate constant at T = ∞ will be equal to frequency factor i.e. A = 6 × 1014 s-1
The rate data for the net reaction at 25°C for the reaction X + 2Y → 3Z are given below :
[X0] [Y0] Time required for [Z] to increase by 0.005 mol per litre.
0.01 0.01 72 sec
0.02 0.005 36 sec
0.02 0.01 18 sec
The intial rate (as given by Z) is :
Rate becomes four times by doubling the concentration of A
⇒ Order w.r.t A is 2
by doubling the concentration of B, the rate becomes double
⇒ Order w.r.t. B is 1.
The rate of production of NH3 in N2 + 3H2 → 2NH3 is 3.4 kg min-1. The rate of consumption of H2 is :
= 5.1 Kgmin-1
For a given reaction of first order it takes 20 min. for the conc. to drop from 1.0 M to 0.60 M. The time required for the conc. to drop from 0.60 M to 0.36 M will be :
Since, Rate constant will be equal
On solving, t = 20 min
For a first order reaction, the concentration of reactant :
For a first order reaction, the concentration of the reactant varies exponentially with time (A = A0e–kt)
Radioactivity of a sample (z = 22) decreases 90% after 10 years. What will be the half-life of the sample?
Reaction is or zero order hence, Option C will be correct.
Mathematical representation for t1/4 life for first order reaction is over is given by :
t1/4 = time taken for decomposition.
For a reaction A → Products, the conc. of reactant C0, aC0, a2C0, a3C0............ after time interval 0, t, 2t ............ where 'a' is constant. Then :
If we calculate K, it comes constant every time i.e. reaction is of first order.
Half-life is independent of conc. of A. After 10 minutes volume N2 gas is 10 L and after complete reaction 50 L. Hence rate constant in min-1 :
Half life is independent of concentration
⇒ reaction is of first order
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
In acidic medium the rate of reaction between (BrO3)- & Br- ions is given by the expression, –[d(BrO3-) /dt] = K[BrO3-][Br-][H +]2 It means :
D option is correct.