Second Order Reaction
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In a reaction, A+B→ Product, rate is doubled when the concentration of B is doubled and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as :
Rate =k[A][B]2
Rate =k[A]2[B]2
Rate =k[A][B]
Rate =k[A]2[B]
Run | [A] mol L−1 | [B] mol L−1 | Initial rate of formation of D mol L−1 min−1 |
I. | 0.1 | 0.1 | 6.0×10−3 |
II. | 0.3 | 0.2 | 7.2×10−2 |
III. | 0.3 | 0.4 | 2.88×10−1 |
IV. | 0.4 | 0.1 | 2.40×10−2 |
(Where, [A] and [B] are inital concentration of A and B respectively.)
- Rate=k[A]2[B]
- Rate=k[A][B]
- Rate=k[A]2[B]2
- Rate=k[A][B]2
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reaction is
Doubled
Reduced to half?
- 2
- 4
- 12
- 14
- 34.07 seconds
- 61.09 seconds
- 24.07 seconds
- 14.09 seconds
Here,
t12 is the half life period.
(Concentration)i is the intial concentration of the reactant.
Explain half-life for zero-order and second-order reactions
A→products
The slope of the graph is found out to be −2
The time (t) is taken in minutes.
The value of rate constant is :
- 2.303 min−1
- 4.606 min−1
- 0.3010 min−1
- 2 min−1
Where the rate law is defined as: −Δ[A]Δt=k[A]2[B][C]
An experiment is carried out where [B]o=[C]o=1.00 M and [A]o=1.00×10−4 M, Calculate the value of t12.
Here,
t12 is the half life period for first order reaction
k is the first order rate constant
- The graph between t12 (y-axis) and initial concentration of reactant (x-axis) is a straight line parallel to x-axis
- The graph between t12 y-axis) and initial concentration of reactant (x-axis) is a rectangular hyperbola
- t12=ln2k
- t12=[A]02k
3A+2B→C+D, the experimental rate law is R=K[A]2[B]1. When the active mass of 'B' is kept constant and 'A' is tripled then, the rate of the reaction will
- Decrease by 3 times
- Increase by 9 times
- Increase by 3 times
- Unpredictable
- Instantaneous rate
- Average rate
- Non-spontaneous rate
- Spontaneous rate
- 2×10−3M/min
- 10−3M/min
- 10−4M/min
- 5×10−4M/min
- rinst.=limΔt→0 (ravg.)
- ravg.=limΔt→0 (rinst.)
- ravg.=limΔC→0 (rinst.)
- rinst.=limΔC→0 (ravg.)
A→Products
Which of the following represents the correct graph between [A] v/s t
- It is proportional to initial concentration for zeroth order
- Average life=1.44×half-life; for first order reaction
- Time of 75% reaction is thrice of half-life period in first order reaction
- 99.9% reaction takes place in 100 minutes for the case when rate constant is 0.0693 min−1
Explain reaction rate with definition.
- None of the above
- None of the above
- 5
- 4
- 2
- 3
2N2O5→4NO2(g)+O2(g)
- The concentration of the reactant decreases exponentially with time
- The reaction proceeds to 99.6% completion in eight half-life duration
- The half-life of the reaction depends on the initial concentration of the reactant
- The half-life of the reaction decreases with increasing temperature
- Second
- Zero
- First
- Third
In a reaction, A+B→ Product, rate is doubled when the concentration of B is doubled and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as :
Rate =k[A][B]2
Rate =k[A]2[B]2
Rate =k[A][B]
Rate =k[A]2[B]
A+2B→C+D
Experiment | Initial concentration [A] |
Initial concentration [B] |
Initial rate of formation of D |
1 | 0.1 | 0.1 | 6×10−3 |
2 | 0.3 | 0.2 | 7×10−2 |
3 | 0.3 | 0.4 | 2.8×10−1 |
4 | 0.4 | 0.1 | 2.4×10−2 |
What will be correct rate law expression for the given reaction?
- Rate=k[A][B]
- Rate=k[A][B]2
- Rate=k[A]2[B]2
- Rate=k[A]2[B]
- 2 hr
- 0.5 hr
- 0.25 hr
- 1 hr
(I) dxdt=K[x][y] (II) dxdt=K[x][y]2
(III) dxdt=k[x]2 (IV) dxdt=K[x]+K[y]2
- I only
- I and III only
- I and II only
- I and IV only
- None of the above
- pressure
- concentration of Reactant
- concentration of product
- temparature