Advertisements
Advertisements
प्रश्न
Consider a certain reaction \[\ce{A -> Products}\] with k = 2.0 × 10−2 s−1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L−1.
Advertisements
उत्तर १
Given: k = 2.0 × 10−2 s−1
T1 = 0
[A]1 = 1.0 M
T2 = 100 s
[A]2 = ?
k = `1/(T_2 - T_1) ln ([A]_1)/([A]_2)`
2.0 × 10−2 = `1/(100 - 0) ln 1.0/([A]_2)`
(2.0 × 10−2) × 100 = `ln 1.0/([A]_2)`
2 = `ln 1/([A]_2)`
7.389 = `1/([A]_2)`
∴ [A]2 = `1/7.389`
= 0.135 mol L−1
उत्तर २
The units of k (s−1) indicate that the reaction is of first order. For a first-order reaction,
k = `2.303/t log_10 [A]_0/([A])`
∴ 2.0 × 10−2 = `2.303/100 log_10 1.0/([A])`
2.0 × 10−2 = `-2.303/100 log_10 [A]`
⇒ log10 [A] = `-(2.0 xx 10^-2 xx 100)/2.303`
= `-2/2.303`
⇒ log10 [A] = −0.8684
⇒ [A] = antilog10 (−0.8684)
= 0.1354 mol L−1
संबंधित प्रश्न
The rate constant of a first order reaction increases from 4 × 10−2 to 8 × 10−2 when the temperature changes from 27°C to 37°C. Calculate the energy of activation (Ea). (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
What will be the effect of temperature on rate constant?
The activation energy for the reaction \[\ce{2 HI_{(g)} -> H2_{(g)} + I2_{(g)}}\] is 209.5 kJ mol−1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
The decomposition of hydrocarbon follows the equation
k = `(4.5 xx 10^11 s^-1) e^(-28000 K//T)`
Calculate Ea.
Calculate activation energy for a reaction of which rate constant becomes four times when temperature changes from 30 °C to 50 °C. (Given R = 8.314 JK−1 mol−1).
What is the effect of adding a catalyst on Activation energy (Ea)
Explain the following terms :
Half life period of a reaction (t1/2)
Write a condition under which a bimolecular reaction is kinetically first order. Give an example of such a reaction. (Given : log2 = 0.3010,log 3 = 0.4771, log5 = 0.6990).
The rate of chemical reaction becomes double for every 10° rise in temperature because of ____________.
Mark the incorrect statements:
(i) Catalyst provides an alternative pathway to reaction mechanism.
(ii) Catalyst raises the activation energy.
(iii) Catalyst lowers the activation energy.
(iv) Catalyst alters enthalpy change of the reaction.
Why does the rate of a reaction increase with rise in temperature?
For an endothermic reaction energy of activation is Ea and enthalpy of reaction ΔH (both of there in KJ moI–1) minimum value of Ea will be ______.
The activation energy in a chemical reaction is defined as ______.
The slope of Arrhenius Plot `("In" "k" "v"//"s" 1/"T")` of first-order reaction is −5 × 103 K. The value of Ea of the reaction is. Choose the correct option for your answer. [Given R = 8.314 JK−1mol−1]
The activation energy of one of the reactions in a biochemical process is 532611 J mol–1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300 = x × 10–3 k310. The value of x is ______.
[Given: ln 10 = 2.3, R = 8.3 J K–1 mol–1]
A schematic plot of ln Keq versus inverse of temperature for a reaction is shown below
The reaction must be:
A first-order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (Ea) for the reaction. [R = 8.314 J K−1 mol−1]
[Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]
What happens to the rate constant k and activation energy Ea as the temperature of a chemical reaction is increased? Justify.
It is generally observed that the rate of a chemical reaction becomes double with every 10°C rise in temperature. If the generalisation holds true for a reaction in the temperature range of 298 K to 308 K, what would be the value of activation energy (Ea) for the reaction?
