1

### JEE Main 2019 (Online) 10th January Evening Slot

The process with negative entropy change is
A
Dissociation of CaSO4 (s) to CaO(s) and SO3(g)
B
Dissolution of iodine in water
C
Synthesis of ammonia from N2 and H2
D
Sublimation of dry ice

## Explanation

N2(g) + 3H2(g) $\rightleftharpoons$ 2NH3(g); $\Delta$ng < 0
2

### JEE Main 2019 (Online) 10th January Evening Slot

An ideal gas undergoes isothermal compression from 5m3 to 1 m3 against a constant external pressure of 4 Nm–2. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol–1 K–1, the temperature of Al increases by
A
${2 \over 3}K$
B
${3 \over 2}K$
C
1 K
D
2 K

## Explanation

Work done on isothermal irreversible for ideal gas

= $-$Pext (V2 $-$ V1)

= $-$4 N/m2 (1 m3 $-$ 5m3)

= 16 Nm

Isothermal process for ideal gas

$\Delta$U = 0

q = $-$ w

= $-$ 16 Nm

= $-$ 16 J

Heat used to increase temperature of

A$\ell$ q = n Cm $\Delta$T

16 J = 1 $\times$ 24 ${J \over {mol.K}}$ $\times$ $\Delta$T

$\Delta$T = ${2 \over 3}$K
3

### JEE Main 2019 (Online) 11th January Morning Slot

For the chemical reaction X $\rightleftharpoons$ Y, the standard reaction Gibbs energy depends on temperature T (in K) as $\Delta$rGo (in kJ mol–1) = 120 $- {3 \over 8}$ T.

The major component of the reaction mixture at T is :
A
Y if T = 300 K
B
Y if T = 280 K
C
X if T = 350 K
D
X if T = 315 K

## Explanation

X $\rightleftharpoons$ Y

Keq = ${{\left[ Y \right]} \over {\left[ X \right]}}$

If Keq > 1 $\Rightarrow$ [Y] > [X]

and Keq < 1 $\Rightarrow$ [Y] < [X]

We know, $\Delta$Go = -RT ln(Keq)

So when Keq > 1 then $\Delta$Go < 0 and Y is major.

And when Keq < 1 then $\Delta$Go > 0 and X is major.