Thermodynamics

JEE Chemistry · 103 questions · Page 11 of 11 · Click an option or "Show Solution" to reveal answer

Q101

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

Correct Answer

Option B

Solution

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

Q102

For a reaction, A(g)

\to

\ell

);

\Delta

-

3RT. The correct statement for the reaction is :

\Delta

H =

\Delta

\ne

\Delta

H =

\Delta

U = O

\left| {} \right.

\Delta

\left| {} \right.

\left| {} \right.

\Delta

\left| {} \right.

\left| {} \right.

\Delta

\left| {} \right.

\left| {} \right.

\Delta

\left| {} \right.

Correct Answer

Option D

Solution

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|

Q103

\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

Correct Answer

Option A

Solution

l

)

\overset{\,}\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

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