Advertisements
Advertisements
Question
Consider the reaction
`3I_((aq))^-) +S_2O_8^(2-)->I_(3(aq))^-) + 2S_2O_4^(2-)`
At particular time t, `(d[SO_4^(2-)])/dt=2.2xx10^(-2)"M/s"`
What are the values of the following at the same time?
a. `-(d[I^-])/dt`
b. `-(d[S_2O_8^(2-)])/dt`
c. `-(d[I_3^-])/dt`
Advertisements
Solution
Rate of reaction
`-(1/3)(d[I^-])/dt = -(d[S_2O_8^-2])/dt = +(d[I_3^-])/dt = +(1/2)(d[SO_4^-2])/dt`
`(d[SO_4^-2])/dt=2.2xx10^-2 "M/s"`
(a)
`-(1/3)(d[I^-])/dt = (1/2)(d[SO_4^-2])/dt`
`-(d[I^-])/dt=3/2xx2.2xx10^-2`
`-(d[I^-])/dt=3.3xx10^-2 " M/s"`
(b)
`-(d[S_2O_s^(-2)])/dt=(1/2)(d[SO_4^-2])/dt`
`-(d[S_2O_s^(-2)])/dt=1/2xx2.2xx10^-2`
`-(d[S_2O_s^(-2)])/dt=1.1xx10^-2 " M/s"`
(c)
`(d[I_3^-])/dt=(1/2)(d[SO_4^-2])/dt`
`(d[I_3^-])/dt=1/2xx2.2xx10^-2`
`(d[I_3^-])/dt=1.1xx10^-2`
`-(d[I_3^-])/dt=-1.1xx10^-2 " M/s"`
APPEARS IN
RELATED QUESTIONS
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
`logk=logA-E_a/2.303R(1/T)`
Where Ea is the activation energy. When a graph is plotted for `logk Vs. 1/T` a straight line with a slope of −4250 K is obtained. Calculate ‘Ea’ for the reaction.(R = 8.314 JK−1 mol−1)
The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?
The decomposition of A into product has value of k as 4.5 × 103 s−1 at 10°C and energy of activation 60 kJ mol−1. At what temperature would k be 1.5 × 104 s−1?
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
In the Arrhenius equation for a first order reaction, the values of ‘A’ of ‘Ea’ are 4 × 1013 sec−1 and 98.6 kJ mol−1 respectively. At what temperature will its half life period be 10 minutes?
[R = 8.314 J K−1 mol−1]
Define activation energy.
What is the effect of adding a catalyst on Activation energy (Ea)
A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given : log 2 = 0·3010, log 4 = 0·6021, R = 8·314 JK–1 mol–1)
The decomposition of a hydrocarbon has value of rate constant as 2.5×104s-1 At 27° what temperature would rate constant be 7.5×104 × 3 s-1if energy of activation is 19.147 × 103 J mol-1 ?
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).
Predict the main product of the following reactions:
The chemical reaction in which reactants require high amount of activation energy are generally ____________.
The rate of chemical reaction becomes double for every 10° rise in temperature because of ____________.
Activation energy of a chemical reaction can be determined by ______.
Consider figure and mark the correct option.
During decomposition of an activated complex:
(i) energy is always released
(ii) energy is always absorbed
(iii) energy does not change
(iv) reactants may be formed
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.
Total number of vibrational degrees of freedom present in CO2 molecule is
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 ______.
In respect of the eqn k = \[\ce{Ae^{{-E_a}/{RT}}}\] in chemical kinetics, which one of the following statement is correct?
The activation energy in a chemical reaction is defined as ______.
Explain how and why will the rate of reaction for a given reaction be affected when the temperature at which the reaction was taking place is decreased.
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]
The equation k = `(6.5 xx 10^12 "s"^(-1))"e"^(- 26000 " K"//"T")` is followed for the decomposition of compound A. The activation energy for the reaction is ______ kJ mol-1. (Nearest integer) (Given: R = 8.314 JK-1 mol-1)
A schematic plot of ln Keq versus inverse of temperature for a reaction is shown below

The reaction must be:
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?
The rate of a reaction quadruples when temperature changes from 27°C to 57°C calculate the energy of activation.
(Given: R = 8. 314 J K−1 mol−1, log 4 = 0.6021)
Given below are two statements:
Statement I: The nutrient deficient water bodies lead to eutrophication.
Statement II: Eutrophication leads to decrease in the level of oxygen in the water bodies.
In the light of the above statements, choose the correct answer from the options given below:
