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

• Chemistry (2018) - Solved questions (2018)

At 518° C,  the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s^-1 when 5% had reacted and 0.5 Torr s^-1  when 33% had reacted. The order of the reaction is

Answer:

Explanation:

For the reaction

$CH_{3}CHO(g)$        $\underrightarrow{Decomposes}$                 $CH_{4}+CO$

Let the order of reaction with respect to CH₃CHO is m.

   Its given, r₁=1 torr/sec. when CH₃CHO is 5% reacted i.e 95% unreacted . Similarly r₂= 0.5 torr/sec when CH₃CHO  is 33% reacted i.e. , 67% unreacted.

    Use the formula,   $r\alpha(a-x)^{m}$

where $(a-x)$ =amount unreacted

so,      $\frac{r_{1}}{r_{2}}=\frac{(a-x_{1})^{m}}{(a-x_{2})^{m}}$ or 

                               $\frac{r_{1}}{r_{2}}=\left[\frac{(a-x_{1})^{}}{(a-x_{2})^{}}\right]^{m}$

Now putting the given values

      $\frac{1}{0.5}=( \frac{0.95}{0.67})^{m}$

$\Rightarrow$        $2= (1.41)^{m}$           or           m=  2

An aqueous solution contain unkonwn concentration of Ba^2+. When 50 mL of a 1M solution of Na₂SO₄ is added , BaSO₄ just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO₄ is 1×10^-10. What is the original concentration of Ba²⁺?

Answer:

Explanation:

It is given that the final volume is 500 mL and this final volume has arrived when 50 mL of 1M Na₂SO₄ was added to unknown Ba²⁺solution. 

So, we can interpret the volume of unknown Ba²⁺ solution as 450 mL i.e.

450 mL Ba²⁺  Solution   +     50mL  Na₂SO₄  Solution →   500mL BaSO₄ Solution

From this, we can calculate the

the concentration of SO₄^2- ion in the solution

via

 M₁V₁=M₂V₂

1×50 =M₂ × 500     (as 1 M Na₂SO₄ is taken into consideration)

M₂= $\frac{1}{10}=0.1 M$

Now for just precipitation,

 Ionic product= Solubility product  ( K_SP)

i.e.

     $[Ba^{2+}][SO_{4}^{2-}]=K_{sp}  $ of BaSO₄

  Given K_sp of BaSO₄  - 1× 10^-10

So,    [Ba²⁺] [01]= 1 × 10^-10

      [Ba²⁺] =1 × 10^-9 M

Reminder  This is the concentration of  Ba²⁺ ions in the final solution. Hence, for calculating  the [Ba²⁺] in the original solution

we have to use

M₁V₁=M₂V₂

as M₁× 450 = 10^-9 × 500

So,

          M₁=1.1 × 10^-9 M

An aqueous solution contains 0.10 M H₂S and 0.20 M  HCl. If the equilibrium constants for the formation of HS^- from H₂S is   1.0×10^-7 and that of S^2- from HS^- ions is 1.2×10^-13 then the concentration of S^2- ions in aqueous solution is

Answer:

Explanation:

Given

     [H₂S]=0.10 M

    [HCl]=0.20 M

    [H⁺]=0.20 M

So,

$H_{2}S\rightleftharpoons H^{+}+HS^{-},K_{1}=1.0\times 10^{-7}$

$HS^{-}\leftrightarrows H^{+}+S^{2-},K_{2}=1.2\times 10^{-13}$

It means for,

  $H_{2}S\rightleftharpoons 2H^{+}+S^{2-}$

$K=K_{1}\times K_{2}$

$1.0\times 10^{-7} \times 1.2\times 10^{-13}$

=1.2× 10^-20

Now      $[S^{2-}]=\frac{K\times [H_{2}S]}{[H^{+}]^{2}}$  [according to the final equation]

   =  $\frac{1.2\times 10^{-20}\times 0.1 M}{(0.2M)^{2}}$

  = $\frac{1.2\times 10^{-20}\times 1 \times 10^{-1}M}{4\times 10^{-2}M}$

   = 3× 10^-20M

The combustion of benzene[l]gives CO₂[g] anf H₂O [l] . Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol^-1 at 25° C ; heat of combustion (in kJ mol^-1) of benzene at constant pressure will be (R= 8.314J K^-1mol^-1)

Answer:

Explanation:

Key Idea   Calculate the heat of combustion with the help of following formula

$\triangle H_{P}=\triangle U+\triangle n_{g}RT$

where $\triangle H_{P}$= Heat of combustion at constant pressure

$\triangle U$= Heat at constant volume (It is also called $\triangle E$)

$\triangle n_{g}$ = Change in number of moles (In gaseous state)

R=Gas constant; T= Temperature.

From the equation,

$C_{6}H_{6}(l)$     +    $\frac{15}{2}O_{2}(g)$   $\rightarrow$    $6CO_{2}(g)$  +   $3H_{2}O(l)$

Change in the number of gaseous moles i.e.

$\triangle n_{g}=6-\frac{15}{2}=-\frac{3}{2} or-1.5$

Now we have $\triangle n_{g}$ and other values given in the question are

$\triangle U=-3263.9kJ/mol$

  T=25°C =273+25=298K

R=8.314JK^-1mol^-1

So, $\triangle H_{p}=(-3263.9)+(-1.5)\times 8.314\times 10^{-3} \times 298$

= -3267.6 kJmol^-1

The trans- alkenes are formed by the reduction of alkynes with

Answer:

Explanation:

Sodium metal in liquid  ammonia reduces alkynes with anti stereochemistry to give trans alkenes.The reduction is selectively anti since the vinyl radical formed during reduction is more stable in trans configuration.

Mechanism

• Chemistry (2018) - Solved questions (2018)