Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry
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221.
Enthalpy of sublimation of a substance is equal to
A
enthalpy of fusion + enthalpy of vapourisation
B
enthalpy of fusion
C
enthalpy of vapourisation
D
twice the enthalpy of vapourisation.
Answer :
enthalpy of fusion + enthalpy of vapourisation
222.
A reaction is spontaneous at low temperature but non-spontaneous at high temperature.
Which of the following is true for the reaction?
A
$$\Delta H > 0,\Delta S > 0$$
B
$$\Delta H < 0,\Delta S > 0$$
C
$$\Delta H > 0,\Delta S = 0$$
D
$$\Delta H < 0,\Delta S < 0$$
Answer :
$$\Delta H < 0,\Delta S < 0$$
We know that $$\Delta G = \Delta H - T\Delta S$$
When $$\Delta H < 0$$ and $$\Delta S < 0$$ then $$\Delta G$$ will be negative at low temperatures ( positive at high temperature ) and the reaction will be
spontaneous.
223.
The values of $$\Delta H$$ and $$\Delta S$$ for the reaction, $${C_{\left( {{\text{graphite}}} \right)}} + C{O_2}\left( g \right) \to 2CO\left( g \right)$$ are $$170\,kJ$$ and $$170\,J{K^{ - 1}},$$ respectively. This reaction will be spontaneous at
A
710$$\,K$$
B
910$$\,K$$
C
1110$$\,K$$
D
510$$\,K$$
Answer :
1110$$\,K$$
$$\eqalign{
& {\text{Given,}}\,\Delta H = 170\,kJ = 170 \times {10^3}J \cr
& \Delta S = 170\,J{K^{ - 1}};\,T = ? \cr
& \Delta G = \Delta H - T\Delta S \cr
& {\text{For spontaneous reaction,}} \cr} $$
$$\Delta G < 0 \Rightarrow 0 < 170 \times {10^3} - T \times 170\,;$$ $$T > 1000$$
$$\therefore \,\,T = 1110\,K$$
224.
If enthalpy of an overall reaction $$X \to Y$$ along one route is $${\Delta _r}H$$ and $${\Delta _r}{H_1},{\Delta _r}{H_2},{\Delta _r}{H_3}\,...$$ representing enthalpies of reactions leading to same product $$Y$$ then $${\Delta _r}H$$ is
225.
The molar heat capacity of water at constant pressure, $${C_p}$$ is $$75\,J\,{K^{ - 1}}mo{l^{ - 1}}.$$ When $$10\,kJ$$ of heat is supplied to $$1\,kg$$ water which is free to expand, the increase in temperature of water is
226.
Two litres of an ideal gas at a pressure of $$10$$ $$atm$$ expands isothermally into a vacuum until its total volume is $$10$$ litres. How much heat is absorbed and how much work is done in the expansion ?
A
$$10\,J,10\,J$$
B
$$8\,J,10\,J$$
C
$$18\,J,0\,J$$
D
$$0\,J,0\,J$$
Answer :
$$0\,J,0\,J$$
$$q = - w = {P_{ex}}\left( {10 - 2} \right) = 0\left( 8 \right) = 0$$
For isothermal expansion in vacuum, $${P_{ex}} = 0$$
227.
The volume of gas is reduced to half from its original volume. The specific heat will ____________.
A
be reduced to half
B
be doubled
C
remain constant
D
be increased four times
Answer :
remain constant
The specific heat is an intensive property which does not depend on the quantity or size of matter.
228.
An ideal gas does work on its surroundings when it expands by $$2.5\,L$$ against external pressure $$2\,atm.$$ This work done is used to heat up $$1\,mole$$ of water at $$293\,K.$$ What would be the final temperature of water in Kelvin if specific heat for water is $$4.184\,J\,{g^{ - 1}}{K^{ - 1}}?$$
A
300
B
600
C
200
D
1000
Answer :
300
Work done,
$$\eqalign{
& w = - {P_{ext.}}\,\,dV \cr
& w = - 2 \times 2.5 \cr
& \,\,\,\,\,\, = - \,5\,L\,atm \cr
& \,\,\,\,\,\, = - 506.3\,J \cr} $$
Because this work is used in raising the temperature of water, so work done is equal to the heat supplied i.e.,
$$w = q = m \cdot {c_s} \cdot \Delta T$$
Given that, $$m = 18\,g\left( { = 1\,mole} \right),{c_s} = 4.184\,J\,{g^{ - 1}}\,{K^{ - 1}},$$ $$q = + 506.3\,J\,$$ ( Heat is given to water ), $$\Delta T = ?$$
$$\eqalign{
& \Delta T = \frac{q}{{{c_s} \cdot m}} \cr
& \,\,\,\,\,\,\,\,\, = \frac{{506.3}}{{4.184 \times 18}} \cr
& \,\,\,\,\,\,\,\,\, = 6.72 \cr} $$
∴ Final temperature,
$$\eqalign{
& {T_f} = {T_i} + \Delta T \cr
& \,\,\,\,\,\,\,\, = 293 + 6.72 \cr
& \,\,\,\,\,\,\,\, = 299.72\,K \approx 300\,K \cr} $$
229.
$$\Delta H$$ for the reaction, $$O{F_2} + {H_2}O \to {O_2} + 2HF$$ ( $$B.E.$$ of $$O - F,O - H,H - F$$ and $$O = O$$ are $$44,111,135$$ and $$119\,kcal\,mo{l^{ - 1}}$$ respectively ) is
230.
Two moles of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litre at 300 $$K.$$ The enthalpy change ( in $$kJ$$ ) for the process is
A
$$11.4 kJ$$
B
$$-11.4 kJ$$
C
$$0 kJ$$
D
$$4.8 kJ$$
Answer :
$$0 kJ$$
$$\Delta H = n{C_p}\,\Delta T$$ solution; since $$\Delta T = 0\,{\text{so}},\,\,\Delta H = 0$$