VITEEE 2016 :: Chemistry :: General questions

• Chemistry - General questions

The following data were  obtained for a given reaction at 300K

 Reaction                             Energy of activation(kJ mol^-1)

 (i) uncatalysed                             76

  (ii)  catalysed                              57

 the factor by which rate of catalysed reaction is increased , is 

Answer:

Explanation:

Using Arrhenius equation

 $K= A.e^{-\frac{E_{a}}{Rt}}$

 $\log k=\log A-\frac{E_{a}}{2.303 RT}$

$\log k_{1}=\log A-\frac{E_{a(1)}{}}{2.303 RT}$    ...............(i)

and $\log k_{2}=\log A-\frac{E_{a(2)}{}}{2.303 RT}$     ...........(ii)

 or     $\log \frac{k_{2}}{k_{1}}=\frac{1}{2.303 RT}[E_{a(1)}-E_{a(2)}]$

                                                             ( from (i) and (ii))

 $=\frac{1}{2.303 \times8.314\times300}[76000-57000]$

 or   $=\log\frac{k_{2}}{k_{1}}=\frac{19000}{2.303\times8.314\times300}$

 $=\frac{190}{6.9\times8.314}$

 or $=\frac{k_{2}}{k_{1}}=2000$    [taking antilog]

A reaction: A₂+B → products, involves the following mechanism

$A_{2}\rightleftharpoons 2A$  (fast)

  (a being the intermediate)

 A+B  $\overrightarrow{k_{2}}$  Products (slow)   The rate law consistent to this mechanism is 

Answer:

Explanation:

 from  slow step :

      rate = k₂ [A][B]  .............(i)

 From fast step :

  $K_{e}=\frac{[A]^{2}}{[A_{2}]} or [A] =  K^{1/2}_{e}[A]^{1/2}$        ...........(ii)

 From (i) and (ii) 

rate =  $k_{2}k_{e}^{1/2}[A_{2}]^{1/2}[B]=k[A_{2}]^{1/2}[B]$

In a reaction, A→ Products, when start is made from 80 x 10^-2 M of A, the half-life is found to be 120 minutes. For the initial concentration  4.0 x 10^-2 M  , the half-life of the reaction becomes 240 minutes. The order of the reaction is 

Answer:

Explanation:

$\frac{(t_{1}/2)_{1}}{(t_{1}/2)_{2}}=(\frac{a_{2}}{a_{2}})^{n-1}$

$\frac{120}{240}=(\frac{4\times 10^{-2}}{8\times 10^{-2}})^{n-1}; n=2$

 A plot In K against $\frac {1}{T}$ (abscissa) is expected to be a straight line with intercept  on ordinate axis equal to 

Answer:

Explanation:

$RT In K =-\triangle G^{0}=T\triangle S^{0}-\triangle H^{0}$;

 $In K =\frac{\triangle S^{0}}{R}-\frac{\triangle H^{0}}{RT}$

 Thus, a plot of  In  K versus 1/T (abscissa)  will be straight line with slope equal to

 $\frac{-\triangle H^{0}}{R}$ and intercept $\frac{\triangle S^{0}}{R}$

A moles of PCl₅ is heated in a closed container to equilibrium   $PCl_{5}\rightleftharpoons PCl_{3}(g)+ Cl_{2}(g)$ at a pressure of p  atm. If x moles of PCl₅ disscociate at equilibrium  , then

Answer:

Explanation:

 $PCl_{5}\rightleftharpoons PCl_{3}(g)+ Cl_{2}(g)$ 

 1-$\alpha$           $\alpha$             $\alpha$

 $\alpha$  = x/a  (degree of dissociation )

 Total moles =  $1-\alpha+\alpha+\alpha=1+\alpha$

$K_{p}=\frac{P_{PCl_{3}}\times P_{Cl_{2}}}{P_{PCl_{5}}}$

       $=\frac{\left[\frac{\alpha}{1+\alpha}.p\right]\left[\frac{\alpha}{1+\alpha}.p\right]}{\frac{1-\alpha}{1+\alpha}.p}$

 $=\frac{\alpha^{2}p}{1-\alpha^{2}}\Rightarrow\alpha=\left(\frac{K_{p}}{K_{p}+P}\right)^{1/2}$

• Chemistry - General questions