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On increasing time, the concentration of  gradually decreasing exponentially.
Given , half-life of radioactive decay = 5730 years So,                  per year we know that, for first-order reaction,                  = 1845 years (approximately) Thus, the age of the sample is 1845 years
The half-life for the first-order reaction is                                                                               = 0.693/4                                                                             =  0.173 year (approximately)
the half-life for the first-order reaction is expressed as ;           =  0.693/2           = 0.35 min (approx)
We know that, half-life () for first-order reaction =                                                             =
The given reaction is first order wrt A and zero order in wrt B. So, the rate of reaction can be expressed as;   Rate = k[A] from exp 1, k = 0.2 per min. from experiment 2nd, [A] =  from experiment 3rd,             from the experiment 4th, from here [A] = 0.1 mol/L
Let assume the rate of reaction wrt A is  and wrt B is  . So, the rate of reaction is expressed as- Rate =  According to given data,  these are the equation 1, 2, 3 and 4 respectively Now,  divide the equation(iv) by (i) we get, from here we calculate that  Again, divide equation (iii) by (ii) from here we can calculate the value of y is 2 Thus, the rate law is now,   So,         ...
we know that  rate law () =  As per data   these are the equation 1, 2 and 3 respectively Now, divide eq.1 by equation2, we get from here we calculate that y = 0 Again, divide eq. 2 by Eq. 3, we get Since y =0 also substitute the value of y  So,   =  =  taking log both side we get,     = 1.496     = approx 1.5 Hence the order of reaction w.r.t A is 1.5 and w.r.t B is 0(zero)
The average rate of reaction between the time 30 s to 60 s is expressed as-
If the concentration of [A] and[B] is increased by 2 times, then                 Therefore, the rate of reaction will increase 8 times.
If the concentration of [B] is increased by 3 times, then                 Therefore, the rate of reaction will increase 9 times.
the reaction is first order in A and second order in B. it means the power of A is one and power of B is 2 The differential rate equation will be-
The rate constant is nearly double when there is a 10-degree rise in temperature in a chemical reaction. effect of temperature on rate constant be represented quantitatively by Arrhenius equation,  where k is rate constant                                                                                                            A is Arrhenius factor           R is gas  constant                 ...
Let assume the concentration of reactant be x So, rate of reaction, R =  Now, if the concentration of reactant is doubled then  . So the rate of reaction would  be   Hence we can say that the rate of reaction reduced to 1/4 times.
Let assume the concentration of reactant be x So, rate of reaction,   Now, if the concentration of reactant is doubled then  . So the rate of reaction would  be   Hence we can say that the rate of reaction increased by 4 times.
The following factors that affect the rate of reaction- the concentration of  reactants  temperature, and presence of catalyst
Given that  So, the unit of rate is bar/min.() And thus the unit of k = unit of rate
The decomposition of  on the platinum surface reaction  therefore,    Rate =  For zero order reaction rate = k therefore,  So    and the rate of production of dihydrogen  = 3(2.5)                                                                             = 7.5
The initial rate of reaction =   substitute the given values of [A], [B] and k, rate  =          =8 When [A] is reduced from 0.1 mol/L to 0.06 mol/L  So, conc. of A reacted = 0.1-0.06 = 0.04 mol/L and conc. of B reacted = 1/2(0.04) = 0.02mol/L conc. of B left = (0.2-0.02) = 0.18 mol/L     Now,  the rate of the reaction is (R) =                                                                  ...
so the order of the reaction is 1 and the dimension of k =
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