1

### JEE Main 2018 (Online) 15th April Evening Slot

Given

(i)   2Fe2O3(s) $\to$ 4Fe(s) + 3O2(g);

$\Delta$rGo = + 1487.0 kJ mol-1

(ii)   2CO(g) + O2(g) $\to$ 2CO2(g);

$\Delta$rGo = $-$ 514.4 kJ mol-1

Free energy change, $\Delta$rGo for the reaction

2Fe2O3(s) + 6CO(g) $\to$ 4Fe(s) + 6CO2(g) will be :
A
$-$ 112.4 kJ mol-1
B
$-$ 56.2 kJ mol-1
C
$-$ 168.2 kJ mol-1
D
$-$ 208.0 kJ mol-1

## Explanation

Given

(i)   2Fe2O3(s) $\to$ 4Fe(s) + 3O2(g);

$\Delta$rGo = + 1487.0 kJ mol-1

(ii)   2CO(g) + O2(g) $\to$ 2CO2(g);

$\Delta$rGo = $-$ 514.4 kJ mol-1

Multiply reaction (ii) with 3 and we get,

(iii)   6CO(g) + 3O2 (g)    $\buildrel \, \over \longrightarrow$    6CO2(g) ;

$\Delta$rGo = 3 $\times$ $-$ 514.4 = $-$ 1543.2 kJ mol$-$1

Now add reaction (i)   with  (iii), we get

2Fe2O3(s) + 6CO(g)    $\buildrel \, \over \longrightarrow$    4Fe(s) + 6CO2(g)

$\therefore\,\,\,$ Free energy change of the reaction,

$\Delta$rGo  =  1487.0 $-$ 1543.2  =  $-$ 56.2 kJ/mol
2

### JEE Main 2018 (Online) 15th April Evening Slot

$\Delta$fGo at 500 K for substance 'S' in liquid state and gaseous state are +100.7 kcl mol-1 and +103 kcal mol-1, respectively. Vapour pressure of liquid 'S' at 500 K is approximately equal to : ( R = 2 cal K-1 mol-1 )
A
0.1 atm
B
1 atm
C
10 atm
D
100 atm

## Explanation

S($l$)   $\buildrel \, \over \longrightarrow$   S(g)

$\Delta$Go = $\Delta$fGo (Vapour) $-$ $\Delta$f Go (liquid)

= 103 $-$ 100.7

= 2.3 kcal / mol

= 2.3 $\times$ 103 cal/mol.

$\Delta$Go = $-$ RT$l$nk

$\Rightarrow $$\,\,\, 2.3 \times 103 = - 2.303 \times 2 \times 500 log K \Rightarrow$$\,\,\,$ log K = $-$ 1

$\Rightarrow$$\,\,\,$ K = 0.1 atm
3

### JEE Main 2018 (Online) 16th April Morning Slot

At 320 K, a gas A2 is 20% dissociated to A(g). The standard free energy change at 320 K and 1 atm in J mol-1 is approximately : (R = 8.314 JK-1 mol-1; ln2 = 0.693; ln 3 = 1.098)
A
4763
B
2068
C
1844
D
4281

## Explanation

A2 (g) $\rightleftharpoons$ 2 A (g)

Assume initially concentration of A2 = [A2] = 1 m

at equilibrium [A2] = 1 $\times$ ${{80} \over {100}}$ = 0.8 M

and 20% of [A2] = 1 $\times$ ${{20} \over {100}}$ = 0.2 M

$\therefore\,\,\,\,$ [A] = 2 $\times$ 0.2 = 0.4 M

Equilibrium constant

K = ${{{{[A]}^2}} \over {[{A_2}]}}$ = ${{{{\left[ {0.4} \right]}^2}} \over {\left[ {0.8} \right]}}$ = 0.2

$\Delta$Go = $-$ RT $\ell$nK

= $-$ 8.314 $\times$ 320 $\times$ $\ell$n(0.2)

= 4281 J/mol.
4

### JEE Main 2018 (Online) 16th April Morning Slot

For which of the following processes, $\Delta$S is negative ?
A
H2(g) $\to$ 2H(g)
B
N2(g, 1 atm) $\to$ N2(g, 5 atm)
C
C(diamond) $\to$ C(graphite)
D
N2(g, 273 K) $\to$ N2(g, 300 K)

## Explanation

N2(g, 1 atom) $\buildrel \, \over \longrightarrow$ N2(g, 5 atom)

Here pressure increases. When pressure increases then the molecules of will come closer and intermoleculer distance decreases, so entropy will also decreases and $\Delta S\, < \,0$.