JEE 2018 :: Chemistry (2018) :: Solved questions (2018)

• Chemistry (2018) - Solved questions (2018)

Consider the following reversible reaction,$ A(g)+B(g)\rightleftharpoons AB (g)$. The activation energy of the backward reaction exceeds that of the forward reaction by 2 RT (in J mol^-1). If the pre-exponential factor of the forward  reaction is times that of the reverse reaction, the absolute value of $\triangle G^{\ominus}$  ( in J mol^-1) for the reaction at 300K is.................                     (Given; ln(2)=0.7 RT =2500J mol^-1 at 300K and G is the Gibbs energy )

Answer:

Explanation:

 For the reaction,

 $ A(g)+B(g)\rightleftharpoons AB (g)$

 Given $E_{a_{b}}=E_{a_{f}}+2RT$

$E_{a_{b}}-E_{a_{f}}=2RT$

 Further, $A_{f}=4A_{b}$      or    $\frac{A_{f}}{A_{b}}=4$

   Now, rate constant for forward reaction,

  $k_{f}=A_{f}e^{\frac{-E_{af}}{RT}}$

 Like wise, rate constant for backward reaction

 $k_{b}=A_{b}e^{\frac{-E_{ab}}{RT}}$

At equillibrium,  Rate of forward reaction = Rate of backward reaction

 ie., $k_{f}=k_{b}$

or   $\frac{k_{f}}{k_{b}}= k_{eq}$

 So,  $ k_{eq}= \frac{A_{f}e^{-E_{\frac{a_{f}}{RT}}}}{A_{b}e^{-E_{\frac{a_{b}}{RT}}}}$

  =   $\frac{A_{f}}{A_{b}}e^{\frac{-(E_{a_{f}}-E_{a_{b})}}{RT}}$

 After putting the given values,  $k_{eq_{}}=4e^{2}$

 (as   $E_{a_{b}}-E_{a_{f}}=2RT$  and       $\frac{A_{f}}{A_{b}}=4$ )

  Now,

Δ G⁰ = -RT ln K_eq

= -2500 ln(4 e²)

= -2500 (ln 4 +ln e² )

= -2500 (1.4 +2)

= -2500 × 3.4

= -8500 J/mol

Absolute value = 8500 J/mol    

The surface of copper gets tarnished by the formation of copper oxide. N₂ gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N₂ gas contains 1 mole % of water vapor as an impurity. The water vapor oxidizes copper as per the reaction is given below

$2Cu (g)  + H_{2}O (g) \rightarrow Cu_{2}O (s)+H_{2}(g)$

  $P_{H_{2}}$   is the minimum partial  pressure of H₂ (in bar) needed to prevent the oxidation at 1250 K. The value of In $P_{H_{2}}$  is........

[Given : total pressure = 1 bar , R  ( universal gas constant ) = 8 JK^-1 mol^-1 , ln(10)=2.30 Cu(s)  and Cu₂ O(s) are  mutually immiscible.)

  At 1250 K

$2 Cu (s)+\frac{1}{2}O_{2}(g)\rightarrow Cu_{2}O (s);$

 $\triangle G^{\ominus} $=-78,000 J mol^-1 

$H_{2}(g)+\frac{1}{2}O_{2}\rightarrow H_{2}O (g);$

$\triangle G^{\ominus}$ =-178,000 J mol^-1

  G  is the Gibs energy

Answer:

Explanation:

Given

(i)  $2 Cu (s)+\frac{1}{2}O_{2}(g)\rightarrow Cu_{2}O (s);$

  Δ G⁰ =-78,000 J mol^-1

  = -78 kJ mol^-1

 (ii)$H_{2}(g)+\frac{1}{2}O_{2}\rightarrow H_{2}O (g);$

  Δ G⁰ =-178,000 J mol^-1

  = -178 kJ mol^-1

   So, net reaction  is (By (i) - (ii))

 $2Cu (g)  + H_{2}O (g) \rightarrow Cu_{2}O (s)+H_{2}(g)$

  Δ G = 100000 J /mol or 10⁵ J/mol=100 kJ mol^-1

  Now, for the above reaction

  $\triangle G = \triangle G^{0}+RT ln[\frac{p_{H_{}}}{p_{H_{2}O}}]$ and to prevent above the reaction.

   Δ G≥ 0

So,   $\triangle G^{0}+RT ln[\frac{p_{H_{}}}{p_{H_{2}O}}] \geq 0$

 After putting the values,

 $10^{5}+8\times1250  ln[\frac{p_{H_{}}}{p_{H_{2}O}}] \geq 0$

or  $10^{5}+10^{4}  ln[\frac{p_{H_{}}}{p_{H_{2}O}}] \geq 0$

 or $10^{4}  (lnp _{H_{2}}-ln p_{H_{2}O})\geq -10^{5}$

or $lnp _{H_{2}} \geq -10 +ln p_{H_{2}O}$  

 or $lnp _{H_{2}} \geq -10 +2.3log(0.01) as  ( $  p _{H_{2}O}=1$ %) ≥ -10-4.6

In the following reactions sequence, the amount of D (in gram) formed from 10 moles of acetophenone is ...........

 (Atomic weights in g mol ^-1 : H=1, C=12, N=14, O=16, Br= 80. The yield (%) corresponding to the product in each step  is given in the  parenthesis)

Answer:

Explanation:

Given,

The products formed are

So , 1.5 mol of 

                  are produced from

10 moles of acetophenone.

      Molar mass of       

 = 240+14+4+72=330

      Hence , amount of 

                                   produced  is 330x1.5  =495 g

To measure the quantity of Mncl₂ dissolved in an aqueous solution , it was completely converted to KMnO₄ using the reaction.

             $MnCl_{2}+K_{2}S_{2}O_{8}+H_{2}O\rightarrow KMnO_{4}+H_{2}SO_{4}+HCl$

  (equation not balanced)

    Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganatic ion disappeared. The quantity of MnCl₂ (in mg) present in the initial solution is........

[Atomic weights in g mol ^-1 : Mn =55, Cl= 35.5]

Answer:

Explanation:

$Mn^{2+}\rightarrow MnO_{4}^{-}+5e^{-}$

 $S_{2}O_{8}^{2-}+2e^{-}\rightarrow 2SO_{4}^{2-}$

$2Mn^{2+}+5S_{2}O_{8}^{2-} \rightarrow 2Mn O_{4}^{-}+10SO_{4}^{2-}$

Also,

      $2MnO_{4}^{-}+5C_{2}O_{4}^{2-} \rightarrow 2Mn^{2+}+10CO_{2}$

Hence,    $2Mn^{2+}\equiv 5C_{2}O_{4}^{2-}$

             $1MnCl_{2}\equiv 2.5H_{2}C_{2}O_{4}$

Oxalic acid taken   =225 mg

                      $\frac{225}{90}=2.5$ millimoles

  Hence, MnCl₂ = 1 millimoles

                        = (55+71)= 126 mg

 Galena (an ore) is  partially oxidised by passing air through it at high temperature. After some time , the passage of air is stopped, but the heating is continued in a closed furnace such that the content undergo self- reduction. The weight (in kg)of Pb produced per kg of O₂  consumed is............  ( Atomic weights in g mol^-1 O= 16, S =32, Pb= 207)

Answer:

Explanation:

The equations of chemical reactions occurring  during the process are-

  In the presence of oxygen

 $2PbS+3O_{2}\rightarrow 2PbO+2SO_{2}.....(i)$

   By self reduction

        $2PbO+PbS_{}\rightarrow 3Pb+SO_{2}$

Thus 3 moles of 0₂ produces 3 moles of Pb 

    i.e.,

  32 × 3=96 g of O₂ produces

  3 × 207 =621 g of Pb

So 1000 g (1kg) of oxygen will produce  $\frac{621}{96}\times 1000= 646815 g$

                                   = 6.4687 kg = 6.47 kg

Alternatively 

       From the direct equation,

        $Pb S+O_{2}\rightarrow Pb +SO_{2}$

                     32g        207g

So,   32g of O₂  gives 207 g of Pb

1 g of O₂ will give  $\frac{207}{32}$ g of Pb

1000g of O₂  will give    $\frac{207}{32}\times 1000$

                            = 6468.75g= 6.46875 kg

                            =6.47 kg

• Chemistry (2018) - Solved questions (2018)