1

JEE Main 2017 (Online) 8th April Morning Slot

The enthalpy change on freezing of 1 mol of water at 5oC to ice at −5oC is :

(Given $\Delta$fusH = 6 kJ mol$-$1 at 0oC,
Cp(H2O, $\ell$ = 75.3J mol$-$1 K$-$1)
Cp(H2O s) =36.8 J mol$-$1 K$-$1)
A
5.44 kJ mol$-$1
B
5.81 kJ mol$-$1
C
6.56 kJ mol$-$1
D
6.00 kJ mol$-$1
2

JEE Main 2017 (Online) 9th April Morning Slot

An ideal gas undergoes isothermal expansion at constant pressure. During the process :
A
enthalpy increases but entropy decreases.
B
enthalpy remains constant but entropy increases.
C
enthalpy decreases but entropy increases.
D
Both enthalpy and entropy remain constant.
3

JEE Main 2017 (Online) 9th April Morning Slot

A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A :
A
10 J of the work will be done by the gas.
B
6 J of the work will be done by the gas.
C
10 J of the work will be done by the surrounding on gas.
D
6 J of the work will be done by the surrounding on gas.
4

JEE Main 2018 (Offline)

The combustion of benzene(l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25oC; heat of combustion (in kJ mol–1) of benzene at constant pressure will be
(R = 8.314 JK–1 mol–1)
A
–3267.6
B
4152.6
C
–452.46
D
3260

Explanation

Formula of Heat of combination is

$\Delta H = \Delta u\,\, + \,\,\Delta ng\,\,RT$

Where, $\Delta H$ $=$ Heat of combination at constant pressure

$\Delta u\, =$ Heat at constant volume

$\Delta {n_g}$ change in number of moles for gaseous molecule.

R = gas constant

T = Temperature.

The Required reaction

C6H6(l) + ${{15} \over 2}$ O2(g) $\to$ 6CO2(g) + 3H2O(l)

Here O2 and CO2 are gaseous molecules so to calculate $\Delta {n_g}$ we only consider those.

$\therefore\,\,\,$ $\Delta {n_g}$ $=$ 6 $-$ ${{15} \over 2}$ = $-$ ${3 \over 2}$

Given $\Delta$u = $-$ 3263.9 kJ mol$-$1 R = 8.314 JK$-$ mol$-$1

R = 8.314 JK$-$1 mol$-$1

= 8.314 $\times$ 10$-$3 kJ K$-$1 mol$-$1

T = 25o C

= 25 + 273 K

= 298 K

So, $\Delta H$ = $-$ 3263.9 + $\left( { - {3 \over 2}} \right)$ $\times$ 8.314 $\times$ 10$-$3 $\times$ 298

= $-$ 3267.6 kJ mol$-$1