Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry
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151.
The enthalpies of formation of $$A{l_2}{O_3}$$ and $$C{r_2}{O_3}$$ are $$ - 1596\,kJ$$ and $$ - 1134\,kJ$$ respectively. $$\Delta H$$ for the reaction $$2Al + C{r_2}{O_3} \to 2Cr + A{l_2}{O_3}$$ is
152.
The heats of neutralisation of $$C{H_3}COOH,HCOOH,HCN$$ and $${H_2}S$$ are $$ - 13.2, - 13.4, - 2.9$$ and $$ - 3.8\,kcal$$ per equivalent respectively. Arrange the acids in increasing order of strength
A
$$HCOOH > C{H_3}COOH > {H_2}S > HCN$$
B
$$C{H_3}COOH > HCOOH > {H_2}S > HCN$$
C
$${H_2}S > HCOOH > C{H_3}COOH > HCN$$
D
$$HCOOH > {H_2}S > C{H_3}COOH > HCN$$
Answer :
$$HCOOH > C{H_3}COOH > {H_2}S > HCN$$
The greater the (negative value) of heat of neutralisation, the more is the strength of the acid. Hence, $$HCOOH > C{H_3}COOH > {H_2}S > HCN$$
153.
Unit of entropy is
A
$$J{K^{ - 1}}\,mo{l^{ - 1}}$$
B
$$J\,mo{l^{ - 1}}$$
C
$${J^{ - 1}}\,{K^{ - 1}}\,mo{l^{ - 1}}$$
D
$$JK\,mo{l^{ - 1}}$$
Answer :
$$J{K^{ - 1}}\,mo{l^{ - 1}}$$
Entropy change equal to change in heat per degree.
$$\Delta S = \frac{q}{T}$$
$$q = $$ required heat per mol
$$T = $$ constant absolute temperature
Thus, unit of entropy is $$J{K^{ - 1}}mo{l^{ - 1}}$$
154.
A gas is allowed to expand in a well insulated container against a constant external pressure of $$2.5\,atm$$ from an initial volume of $$2.50L$$ to a final volume of $$4.50 L.$$ The change in internal energy $$\Delta U$$ of the gas in joules will be
A
$$1136.25\,J$$
B
$$ - 500\,J$$
C
$$ - 505\,J$$
D
$$ + 505\,J$$
Answer :
$$ - 505\,J$$
Key concept According to first law of thermodynamics,
$$\Delta U = q + w$$
where, $$\Delta U = $$ internal energy
$$q =$$ heat absorbed or evolved, $$w =$$ work done.
Also, work done against constant external pressure ( irreversible process ).
$$w = - {p_{ext}}\,\Delta V.$$
Work done in irreversible process,
$$\eqalign{
& w = - {p_{ext}}\,\Delta V = - {p_{ext}}\left( {{V_2} - {V_1}} \right) \cr
& \,\,\,\,\,\, = - 2.5\,atm\,\left( {4.5\,L - 2.5\,L} \right) \cr
& \,\,\,\,\,\, = - 5\,L\,atm = - 5 \times 101.3\,J \cr
& \,\,\,\,\,\, = - 505\,J \cr} $$
Since, the system is well insulated, $$q=0$$
$$\therefore \,\,\Delta U = w = - 505\,J$$
Hence, change in internal energy, $$\Delta U$$ of the gas is $$ - 505\,J.$$
155.
Bond dissociation enthalpies of $${H_{2\left( g \right)}}$$ and $${N_{2\left( g \right)}}$$ are $$436.0\,kJ\,mo{l^{ - 1}}$$ and $$941.8\,kJ\,mo{l^{ - 1}},$$ respectively, and enthalpy of formation of $$N{H_{3\left( g \right)}}$$ is $$ - 46\,kJ\,mo{l^{ - 1}}.$$ What are the enthalpy of atomisation of $$N{H_{3\left( g \right)}}$$ and the average bond enthalpy of $$N - H$$ bond respectively ( in $$kJ\,mo{l^{ - 1}}$$ ) ?
A
1170.9, 390.3
B
117, 300
C
300, 200
D
2000, 1975
Answer :
1170.9, 390.3
$${N_{2\left( g \right)}} + 3{H_{2\left( g \right)}} \to 2N{H_{3\left( g \right)}};$$ $$\Delta H = - 2 \times 46\,kJ\,mo{l^{ - 1}}$$
$$\eqalign{
& \Delta H = \sum {{{\left( {BE} \right)}_{{\text{reactants}}}}} - \sum {{{\left( {BE} \right)}_{{\text{products}}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \left( {941.8 + 3 \times 436} \right) - \left( {6x} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\, = - 2 \times 46 \cr
& {\text{(here }}x = BE{\text{ of }}N - H{\text{ bonds)}} \cr
& x = 390.3\,kJ\,mo{l^{ - 1}} \cr
& N{H_3} \to N + 3\left( H \right) \cr} $$
$${\text{Heat of atomisation}}$$ $$ = 3 \times 390.3 = 1170.9\,kJ\,mo{l^{ - 1}}$$
156.
A gas expands adiabatically at constant pressure such that $$T \propto {V^{ - \,\frac{1}{2}}}.$$ The value of $$\gamma \left( {{C_{p,m}}/{C_{v,m}}} \right)$$ of the gas will be :
157.
The enthalpy of neutralisation of $$N{H_4}OH$$ and $$C{H_3}COOH$$ is $$ - 10.5\,kcal\,mo{l^{ - 1}}$$ and enthalpy of neutralisation of $$C{H_3}COOH$$ with strong base is $$ - 12.5\,kcal\,mo{l^{ - 1}}.$$ The enthalpy of ionisation of $$N{H_4}OH$$ will be
When gas is compressed, its entropy decreases so, $$\Delta S$$ is negative.
160.
The enthalpy of formation of ammonia when calculated from the following bond energy data is ( $$B.E.$$ of $$N - H,$$ $$H-H,$$ $$N \equiv N$$ is $$389\,kJ\,mo{l^{ - 1}},$$ $$435\,kJ\,mo{l^{ - 1}},$$ $$945.36\,kJ\,mo{l^{ - 1}}$$ respectively )