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.
(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 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)
What will be the effect of temperature on rate constant?
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
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 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?
The decomposition of hydrocarbon follows the equation
k = `(4.5 xx 10^11 s^-1) e^(-28000 K//T)`
Calculate Ea.
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.
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)
Explain the following terms :
Half life period of a reaction (t1/2)
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 in the redox titration of \[\ce{KMnO4}\] vs oxalic acid, we heat oxalic acid solution before starting the titration?
What happens to most probable kinetic energy and the energy of activation with increase in temperature?
Total number of vibrational degrees of freedom present in CO2 molecule is
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]
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 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:
Activation energy of any chemical reactions can be calculated if one knows the value of:
