1

### JEE Main 2016 (Online) 10th April Morning Slot

If 100 mole of H2O2 decompose at 1 bar and 300 K, the work done (kJ) by one mole of O2(g) as it expands against 1 bar pressure is :

2H2O2(l)    $\rightleftharpoons$    2H2O(l) + O2(g)

(R = 8.3 J K $-$1 mol$-$1)
A
62.25
B
124.50
C
249.00
D
498.00
2

### JEE Main 2017 (Offline)

$\Delta$U is equal to
A
Isobaric work
B
C
Isothermal work
D
Isochoric work

## Explanation

From 1st law of thermodynamics

ΔU = q + w

q = 0

$\therefore$ ΔU = w

So change in internal energy (ΔU) is equal to adiabatic work.
3

### JEE Main 2017 (Offline)

Given, ${C_{(graphite)}} + {O_2} \to C{O_2}(g)$;
${\Delta _r}{H^o}$ = - 393.5 kJ mol-1
${{\rm H}_2}(g)$ + ${1 \over 2}{O_2}(g)$$\to {{\rm H}_2}{\rm O}(l) {\Delta _r}{H^o} = - 285.8 kJ mol-1 C{O_2}(g) + 2{{\rm H}_2}{\rm O}(l) \to C{H_4}(g) + 2{O_2}(g) {\Delta _r}{H^o} = + 890.3 kJ mol-1 Based on the above thermochemical equations, the value of {\Delta _r}{H^o} at 298 K for the reaction {C_{(graphite)}} + 2{{\rm H}_2}(g) \to C{H_4}(g) will be: A +144.0 kJ mol–1 B – 74.8 kJ mol–1 C -144.0 kJ mol–1 D + 74.8 kJ mol–1 ## Explanation C(graphite) + O2(g) \to CO2 (g); {\Delta _r}{H^o} = - 393.5 kJ mol-1 .............(1) {{\rm H}_2}(g) + {1 \over 2}{O_2}(g)$$\to {{\rm H}_2}{\rm O}(l)$
${\Delta _r}{H^o}$ = - 285.8 kJ mol-1 .........(2)

$C{O_2}(g)$ + $2{{\rm H}_2}{\rm O}(l) \to$ $C{H_4}(g)$ + $2{O_2}(g)$
${\Delta _r}{H^o}$ = + 890.3 kJ mol-1 .........(3)

C(graphite) + 2H2(g) $\to$ CH4 (g);    $\Delta$H = ? ........ (4)

You can see,

[Eq. (1) + Eq. (3)] + [2 × Eq. (2)] = Eq. (4)

$\therefore$ [ $\Delta$rH1 + $\Delta$rH3 ] + [2 $\times$ $\Delta$rH2] = $\Delta$H

$\Rightarrow$ $\Delta$H = [(-393.5) + (890.3)] + [2(-285.8)] = -74.8 kJ / mol
4

### JEE Main 2017 (Online) 8th April Morning Slot

For a reaction, A(g) $\to$ A($\ell$); $\Delta$H= $-$ 3RT.
The correct statement for the reaction is :
A
$\Delta$H = $\Delta$U $\ne$ O
B
$\Delta$H = $\Delta$U = O
C
$\left| {} \right.$$\Delta H\left| {} \right. < \left| {} \right.$$\Delta$U$\left| {} \right.$
D
$\left| {} \right.$$\Delta H\left| {} \right. > \left| {} \right.$$\Delta$U$\left| {} \right.$

## Explanation

Given reaction,

A(g)  $\to$  A($l$) ;

We know,

$\Delta$H = $\Delta$V + $\Delta$ng RT

Given, $\Delta$H = $-$3RT

From reaction,

$\Delta$ng = 0$-$ 1 = $-1$

$\therefore\,\,\,$ $\Delta$H = $\Delta$U $-$ RT

$\Rightarrow$ $\,\,\,$ $-$ 3RT = $\Delta$U $-$ RT

$\Rightarrow$$\,\,\,$ $\Delta$U = $-$ 2RT

$\therefore\,\,\,$ $\left| {\Delta U} \right|$ = 2RT

and $\left| {\Delta H} \right|$ = 3RT

$\therefore\,\,\,$ $\left| {\Delta H} \right|$ > $\left| {\Delta U} \right|$