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4.9     The activation energy for the reaction 2HI(g)\rightarrow H_{2}+I_{2}(g)  is 209.5 KJ mol^{-1} at 518 K.  Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

We have  Activation energy = 209.5KJ/mol temperature= 581K R = 8.314J/mol/K Now,  the fraction of molecules of reactants having energy equal to or greater than activation energy is given as  taking log both sides we get                       = 18.832 x = antilog(18.832)    = 1.471  

4.8     The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K Calculate E_{a}.

Given data  (initial temperature) = 298K and  (final temperature)= 308K And we know that rate of reaction is nearly doubled when temperature rise 10-degree So,  and R = 8.314 J/mol/K now,  On putting the value of given data we get,    Activation energy () =                                        =52.9 KJ/mol(approx)

4.7     What will be the effect of temperature on rate constant ?

The rate constant of the reaction is nearly doubled on rising in 10-degree temperature. Arrhenius equation depicts the relation between temperature and rates constant. A= Arrhenius factor Ea = Activation energy R = gas constant T = temperature

4.6     Time required to decompose SO_{2}Cl_{2} to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

We know that t(half ) for the first order reaction is   and we have given the value of half time  thus,               = 0.01155 /min        OR = 1.1925 Alternative method  we can also solve this problem by using the first order reaction equation.                              put 

4.5     A first order reaction has a rate constant 1.15\times 10^{-3}s^{-1}  . How long will 5g of his reactant take to reduce to 3g?

Given data, initial conc. = 5g final conc. = 3g rate const. for first order =  We know that for the first order reaction,                               [log(5/3)= 0.2219]     = 444.38 sec (approx)      

4.4     The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y ?

The order of a reaction means the sum of the power of concentration of the reactant in rate law expression. So rate law expression for the second-order reaction is  here R = rate if the concentration is increased to 3 times means  new rate law expression =  = =  9R the rate of formation of Y becomes 9 times faster than before                                         

4.3    For a reaction, A+B\rightarrow product; the rate law is given by, r=k\left [ A \right ]^{1/2}\left [ B \right ]^{2}. What is the order of the reaction?

Order of reaction = Sum of power of concentration of the reactant in the rate law expressions So, here power of A = 0.5 and power of B = 2 order of reaction  = 2+0.5 =2.5

4.2  In a reaction,2A\rightarrow Products P, the concentration of A decreases from 0.5 mol L^{-1} to 0.4 mol L^{-1} in 10 minutes. Calculate the rate during this interval?

According to the formula of an average rate                                                                    =             (final concentration - initial conc.)/time interval                                                                   =                                                                    =                                                                    =              ...

.4.1     For the reaction R\rightarrow P, the concentration of a reactant change from 0.03M to 0.02Min 25  minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

We know that, The average rate of reaction =                                                =                                                =                                                =  In seconds we need to divide it by 60. So,                                               =                                                = 6.67 
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