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From the Arrhenius equation, ...................................(i) it is given that  T1= 293 K T2 = 313 K Putting all these values in equation (i) we get, Activation Energy = 52.86 KJ/mol   This is the required activation energy
We know that, for a first order reaction- Case 1 At temp. = 298 K    = 0.1054/k Case 2 At temp = 308 K       = 2.2877/k' As per the question    K'/K = 2.7296 From Arrhenius equation,      = 76640.096 J /mol      =76.64 KJ/mol   k at 318 K we have , T =318K                  A=  Now  After putting the calue of given variable, we get  on takingantilog we get,  k =...

S seema garhwal
The decomposition of A into a product has a value of k as at 10°C and energy of activation 60 kJ mol–1. K1 =  K2 =   =  60 kJ mol–1 K2 =
The Arrhenius equation is given by  taking log on both sides,  ....................(i) given equation, .....................(ii) On comparing both equation we get, activation energy  half life ()  = 256 min k = 0.693/256    With the help of equation (ii),            T =                 = 669 (approx)
The Arrhenius equation is given by  .................................(i) given equation, ............................(ii) by comparing equation (i) & (ii) we get, A= 4.51011 per sec Activation energy = 28000  (R = 8.314)                              = 232.792 KJ/mol
For first order reaction, given that half life = 3 hrs () Therefore k = 0.693/half-life                     = 0.231 per hour Now,                   = antilog (0.8024)                   = 6.3445 (approx) Therefore fraction of sample of sucrose remains after 8 hrs is 0.157
Given that, k =  t = 100 s Here the unit of k is in per sec, it means it is a first-order reaction. therefore,  Hence the concentration of rest test sample is 0.135 mol/L
Given that, k =  = 179.9 KJ/mol T(temp) = 546K According to Arrhenius equation, taking log on both sides,                             = (0.3835 - 5) + 17.2082               = 12.5917 Thus A = antilog (12.5917)          A = 3.9  per sec (approx)
From the above data, T/ 0 20 40 60 80 T/K 273 293     313 333 353 () 3.66 3.41 3.19 3.0 2.83 0.0787 1.70 25.7 178 2140 -7.147 -4.075 -1.359 -0.577 3.063 Slope of line  =  According to Arrhenius equations, Slope =    12.30 8.314             = 102.27  Again, When T = 30 +273 = 303 K and 1/T =0.0033K    k =  When T = 50  + 273 = 323 K  and 1/T = 3.1  K k = 0.607 per sec
The thermal decomposition of  is shown here; After t time, the total pressure  =                                                 So,  thus,  for first order reaction,     now putting the values of pressures, when t = 100s       when      = 0.65 - 0.5    = 0.15 atm So,                              = 0.5 - 0.15                             = 0.35 atm Thus, rate of reaction, when the...
Decompostion is represented by equation- After t time, the total pressure  =                                                 So,  thus,  for first order reaction,     now putting the values of pressures, when t =360sec when t = 270sec So,
For the first-order reaction,               (30% already decomposed and remaining is 70%)      therefore half life = 0.693/k                             =                              = 77.7 (approx)
case 1- for 99% complition,             CASE- II for 90% complition,            Hence proved.
Given, half life = 21.8 years            = 0.693/21.8             and,  by putting the value we get,                            taking antilog on both sides, [R] = antilog(-0.1071)       = 0.781  Thus 0.781  of  will remain after given 10 years of time. Again,  Thus 0.2278  of  will remain after 60 years.
We know that, for first order reaction,           (nearly) Hence the time required is
The half life produce =                                    =
From the log graph, the slope of the graph is =                                            = -k/2.303                             ..(from log equation)        On comparing both the equation we get,
Here, the reaction is in first order reaction because its log graph is linear. Thus rate law can be expessed  as
0 1.63 -1.79 400 1.36 -1.87 800 1.14 -1.94 1200 0.93 -2.03 1600 0.78 -2.11 2000 0.64 -2.19 2400 0.53 -2.28 2800 0.43 -2.37 3200 0.35 -2.46

Half life of the reaction is-
time corresponding to the $1.63 \times 10^{2}/2$ mol/ L  = 81.5 mol /L is the half life of the reaction.  From the graph, the answer shoul be in the range of 1400 s to 1500 s.

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