Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
Explain a graphical method to determine activation energy of a reaction.
The rate constant for the first-order decomposition of H2O2 is given by the following equation:
`logk=14.2-(1.0xx10^4)/TK`
Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes.
(Given: R = 8.314 JK–1 mol–1)
The rate constant of a first order reaction increases from 2 × 10−2 to 4 × 10−2 when the temperature changes from 300 K to 310 K. Calculate the energy of activation (Ea).
(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
The rate constant for the decomposition of N2O5 at various temperatures is given below:
| T/°C | 0 | 20 | 40 | 60 | 80 |
| 105 × k/s−1 | 0.0787 | 1.70 | 25.7 | 178 | 2140 |
Draw a graph between ln k and `1/T` and calculate the values of A and Ea. Predict the rate constant at 30º and 50ºC.
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?
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.
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)
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)
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).
Consider figure and mark the correct option.
Which of the following graphs represents exothermic reaction?
(a)
(b)
(c)
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
The reaction between \[\ce{H2(g)}\] and \[\ce{O2(g)}\] is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of water. Explain.
Why does the rate of a reaction increase with rise in temperature?
Total number of vibrational degrees of freedom present in CO2 molecule is
The rate constant for a reaction is 1.5 × 10–7 sec–1 at 50°C. What is the value of activation energy?
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]
Arrhenius equation can be represented graphically as follows:
The (i) intercept and (ii) slope of the graph are:
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]
A schematic plot of ln Keq versus inverse of temperature for a reaction is shown below
The reaction must be:
What happens to the rate constant k and activation energy Ea as the temperature of a chemical reaction is increased? Justify.
Activation energy of any chemical reactions can be calculated if one knows the value of:
