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Two reactions R₁ and  R₂  have identical pre-exponential factors. The activation energy of R₁ exceeds that of R₂ by 10 kj mol^-1 . If k₁ and k₂ are rate constants for reactions R₁ and R2, respectively at 300 K. then ln $(\frac{k_{2}}{k_{1}})$ is equal to   (R=8.314 J mol^-1 K^-1)

Answer:

Explanation:

According to the Arrhenius equation

                   K= Ae^-E_e/RT 

   where A= collision number or pre-exponentail factor

     R= gas constant

      T= absolute temperature

     E_a = energy of activation

    For reaction R₁ ,k = Ae^-E_a1/RT           ...........(i)

    For reaction R₂ ,k₂ =  Ae^-E_a2/RT   ............(ii)

on dividing Eq  (ii) by Eq (i) , we get

          $\frac{k_{2}}{k_{1}}=e^-{\frac{(E_{a_{2}}-E_{a_{1}})}{RT}}$                .........(iii)

($\because$ Pre -exponental factor 'A' is same for both reactions)

      Taking In on both the sides of Eq  (iii) 

    We get

             $ln(\frac{k_{2}}{k_{1}})=\frac{E_{a_{1}}-E_{a_{2}}}{RT}$

     Given,   $E_{a_{1}}=E_{a_{2}}+10 kJ mol^{-1}$

                                       $E_{a_{1}}=E_{a_{2}}+10000 J mol^{-1}$

   $\therefore ln\frac{k_{2}}{k_{1}}=\frac{10000 J mol^{-1}}{8.314 Jmol^{-1}K^{-1}\times 300K}=4$

3-methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product . The number of possible stereoisomers for the product is

Answer:

Explanation:

The number of stereoisomers in molecules which are not divisble into two equal halves and have n number of asymmetric  C-atoms = 2ⁿ

3-methyl-pen-2-ene on reaction with HBr in presence of peroxide forms an addition product  i.e,   2-bromo-3-methyl pentane. It has two chiral centres.   Therefore 4 stereoisomers are possible.

Both lithium and magnesium display several similar properties due to the diagonal relationship:however the one which is incorrect is

Answer:

Explanation:

Mg can form basic carbonate while Li cannot

              $5 Mg^{2+}+6CO^{2-}_{3}+7 H_{2}O\rightarrow 4MgCO_{3}+Mg(OH)_{2}.5H_{2}O +2HCO_{3}^{-}$

The most abundant elements by mass in the body of a healthy human adult are Oxygen (61.4%); Carbon (22.9%); Hydrogen (10.0%); and Nitrogen (2.6%). The weight which 75 kg person would gain if all  ¹H atoms are replaced by ²H atoms is

Answer:

Explanation:

Given, the abundance of elements of mass 

       oxygen =61.4%,   carbon =22.9% hydrogen = 10.0%, nitrogen = 2.6%

Total weight of person =75 kg

              Mass due to ¹H = $\frac{75\times 10}{100}=7.5$ kg

  ¹H   atoms are replaced by ²H atoms.

     Mass due to ²H = (7.5 × 2)kg

   $\therefore$     Mass gain by person = 7.5 kg

pK_a of a weak acid (HA) and pK_b of a weak base (BOH) are 3.2 and 3.4  respectively. The pH of their salt (AB) solution is

Answer:

Explanation:

For a salt of the weak acid and weak base.

               $pH= 7+\frac{1}{2}pK_{a}-\frac{1}{2}pK_{b}$

Given,    $pK_{a}(HA)= 32,pK_{a}(BOH)=3.4$

    $\therefore$    $pH  =7+\frac{1}{2}(3.2)-\frac{1}{2}(3.4)$

           =7+1.6-1.7 = 6.9    

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