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प्रश्न
Read the following passage and answer the questions that follow:
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The rate of reaction is concerned with decrease in the concentration of reactants or increase in the concentration of products per unit of time. It can be expressed as instantaneous rate at a particular instant of time and average rate over a large interval of time. A number of factors such as temperature, concentration of reactants, catalyst affect the rate of reaction. Mathematical representation of rate of a reaction is given by rate law: Rate = k[A]x [B]y x and y indicate how sensitive the rate is to change in concentration of A and B. Sum of x + y gives the overall order of a reaction. |
- What is the effect of temperature on the rate constant of a reason? [1]
- For a reaction \[\ce{A + B → Product}\], the rate law is given by, Rate = k[A]2 [B]1/2. What is the order of the reaction? [1]
- How order and molecularity are different for complex reactions? [1]
- A first-order reaction has a rate constant 2 × 10–3 s–1. How long will 6 g of this reactant take to reduce to 2 g? [2]
OR
The half-life for radioactive decay of 14C is 6930 years. An archaeological artifact containing wood had only 75% of the 14C found in a living tree. Find the age of the sample.
[log 4 = 0.6021, log 3 = 0.4771, log 2 = 0.3010, log 10 = 1] [2]
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उत्तर
a. The rate constant (k) for a reaction increases with an increase in temperature and becomes almost double with every 10° rise in temperature. This effect is expressed by the Arrhenius equation:
k = `Ae^(-E_a//RT)`
b. According to the equation, r = k[A]2[B]1/2
Order of the reaction = `2 + 1/2 = 5/2`
c. Order of reaction is defined as the sum of the powers of the molar concentration of the reaction species in the rate equation of the reaction. It is applicable to both elementary and complex reactions.
Molecularity of a reaction is defined as the total number of reacting species participating in an elementary reaction. It has no significance for complex reactions as applicable only to elementary reactions.
d. k = 2 × 10–3 s–1
t = `2.303/k log [R]_0/[R]`
t = `2.303/(2 xx 10^-3) log 6/2`
t = 1151.5 × 0.4771
t = 549.3 s
OR
According to first-order reaction,
Half-life (t1/2) = `0.693/k`
t1/2 = 6980 years
k = `0.693/6980`
However, the time taken can be calculated using the first-order rate of reaction when wood contains only 75% of 14C.
Initial concentration of C14, [R]0 = 100
Amount at time t s, [R] = 75
t = `2.303/k log [R]_0/([R])`
t = `2.303/(0.693/6980) log 100/75`
t = 2898 years
Thus, the age of the sample is approximately 2898 years.
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संबंधित प्रश्न
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
From the rate expression for the following reaction, determine the order of reaction and the dimension of the rate constant.
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\[\ce{2A + B → A2B}\]
the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−1. Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1.
A reaction is first order in A and second order in B. How is the rate affected on increasing the concentration of B three times?
Define the following term:
Pseudo first-order reaction
Compounds ‘A’ and ‘B’ react according to the following chemical equation.
\[\ce{A(g) + 2B(g) -> 2C(g)}\]
Concentration of either ‘A’ or ‘B’ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.
| Experiment | Initial concentration of [A]/mol L–¹ |
Initial concentration of [B]/mol L–¹ |
Initial rate of formation of [C]/mol L–¹ s–¹ |
| 1. | 0.30 | 0.30 | 0.10 |
| 2. | 0.30 | 0.60 | 0.40 |
| 3. | 0.60 | 0.30 | 0.20 |
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The role of a catalyst is to change
In the presence of a catalyst, the heat evolved or absorbed during the reaction.
For a reaction 1/2 A ⇒ 2B, rate of disappearance of A is related 't o the appearance of B by the expression:
