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4.9     The activation energy for the reaction   is  at .  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  in absolute temperature from  Calculate .

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  to half of its initial amount is  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   . How long will  of his reactant take to reduce to ?

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, ; the rate law is given by, . 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, P, the concentration of A decreases from to  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 , the concentration of a reactant change from  to in   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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