Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry
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111.
A piston filled with $$0.04\,mol$$ of an ideal gas expands reversibly from $$50.0\,mL$$ to $$375\,mL$$ at a constant temperature of $${37.0^ \circ }C.$$ As it does so, it absorbs $$208\,J$$ of heat. The values of $$q$$ and $$w$$ for the process will be :
$$\left( {R = 8.314\,J/mol\,K} \right)\left( {ln\,7.5 = 2.01} \right)$$
A
$$q = + 208\,J,w = - 208\,J$$
B
$$q = - 208\,J,\,w = - 208\,J$$
C
$$q = - 208\,J,\,w = + 208\,J$$
D
$$q = + 208\,J,\,w = + 208\,J$$
Answer :
$$q = + 208\,J,w = - 208\,J$$
Process is isothermal reversible expansion, hence
$$\eqalign{
& \Delta U = 0,{\text{therefore}}\,\,q = - w. \cr
& {\text{Since}}\,\,q = + 208\,J,\,W = - 208\,J \cr} $$
112.
The enthalpies of combustion of carbon and carbon monoxide are $$- 393.5$$ and $$ - 283\,kJ\,mo{l^{ - 1}}$$ respectively. The enthalpy of formation of carbon monoxide per mole is
A
$$-676.5 kJ$$
B
$$676.5 kJ$$
C
$$110.5 kJ$$
D
$$-110.5 kJ$$
Answer :
$$-110.5 kJ$$
$$\left( {\text{i}} \right)\,C + {O_2} \rightleftharpoons C{O_2},$$ $$\Delta H = - 393.5\,kJmo{l^{ - 1}}$$
$$\left( {{\text{ii}}} \right)\,CO + \frac{1}{2}{O_2} \rightleftharpoons C{O_2},$$ $$\Delta H = - 283.0\,kJmo{l^{ - 1}}$$
$${\text{Operating (i) - (ii), we have }}$$
$$C + \frac{1}{2}{O_2} \to CO\,\,\Delta H = $$ $$ - 110.5\,\,kJmo{l^{ - 1}}$$
113.
$$\Delta G$$ in $$A{g_2}O \to 2\,Ag + \frac{1}{2}{O_2}$$ at a certain temperature is $$ - 10\,kJ\,mo{l^{ - 1}}.$$ Pick the correct statement
A
$$A{g_2}O$$ decomposes to $$Ag$$ and $${O_2}$$
B
$$Ag$$ and $${O_2}$$ combines to form $$A{g_2}O$$
C
Reaction is in equilibrium
D
Reaction does not take place
Answer :
$$A{g_2}O$$ decomposes to $$Ag$$ and $${O_2}$$
$$\Delta G = - ve\,{\text{means the process is spontaneous}}{\text{.}}$$
114.
$${\text{The}}\,\Delta H_f^0\,{\text{for}}\,C{O_2}\left( g \right),\,CO\left( g \right)\,$$ and $${H_2}O\left( g \right)$$ are $$-393.5,$$ $$-110.5$$ and $$ - 241.8\,kJ\,mo{l^{ - 1}}$$ respectively. The standard enthalpy change ( in $$kJ$$ ) for the reaction $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$
A
524.1
B
41.2
C
-262.5
D
-41.2
Answer :
41.2
$$\eqalign{
& {\text{C}}{{\text{O}}_{2\left( g \right)}} + {H_{2\left( g \right)}} \to C{O_{\left( g \right)}} + {H_2}{O_{\left( g \right)}}\,\,,\,\,\Delta H = ? \cr
& \Delta H = \sum {\Delta {H_f}\left( {product} \right) - \sum {\Delta {H_f}\left( {{\text{reactant}}} \right)} } \cr
& {\text{Given,}}\,\,\Delta {H_f}C{O_{2\left( g \right)}} = - 393.5\,kJ/mol \cr
& \Delta {H_f}C{O_{\left( g \right)}} = - 110.5\,kJ/mol \cr
& \Delta {H_f}{H_2}{O_{2\left( g \right)}} = - 241.8\,kJ/mol \cr
& \therefore \,\,\,\Delta H = \left[ {\Delta {H_f}C{O_{\left( g \right)}} + \Delta {H_f}{H_2}{O_{\left( g \right)}}} \right] - \left[ {\Delta {H_f}C{O_{2\left( g \right)}} + \Delta {H_f}{H_{2\left( g \right)}}} \right] \cr
& = \left[ { - 110.5 + \left( { - 241.8} \right)} \right] - \left( { - 393.5 + 0} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\,\Delta {H_f}{{\left( {{H_2}} \right)}_g} = 0} \right] \cr
& = 41.2\,kJ\,mo{l^{ - 1}} \cr} $$
115.
Study the given graph and choose the correct option.
A
$$\Delta H = $$ Net heat absorbed from the surroundings
B
$$\Delta H = $$ Net heat given to the surroundings
C
$$\Delta H = + ve$$ for the reaction
D
$$\Delta H = $$ Total energy possessed by the reactants
Answer :
$$\Delta H = $$ Net heat given to the surroundings
It shows that the reaction is exothermic and $$\Delta H$$ is the heat given to the surroundings.
116.
Thermodynamics is not concerned about __________.
A
energy changes involved in a chemical reaction
B
the extent to which a chemical reaction proceeds
C
the rate at which a reaction proceeds
D
the feasibility of a chemical reaction
Answer :
the rate at which a reaction proceeds
Thermodynamicsis not concerned about how and at what rate chemical reactions are carried out, but is based on initial and final states of a system undergoing the change
117.
The change in energy on freezing $$1.0\,kg$$ of liquid water at $${0^ \circ }C$$ and $$atm$$ is ( Given energy of 1 fusion of ice $$ = 6.03\,kJ\,mo{l^{ - 1}}$$ at $${0^ \circ }C$$ )
A
$$235\,kJ\,k{g^{ - 1}}$$
B
$$450\,kJ\,k{g^{ - 1}}$$
C
$$ - 335\,kJ\,k{g^{ - 1}}$$
D
$$ - 235\,kJ\,k{g^{ - 1}}$$
Answer :
$$ - 335\,kJ\,k{g^{ - 1}}$$
Enthalpy of freezing of water $$ = - 6.03\,kJ\,mo{l^{ - 1}}$$ at $${0^ \circ }C$$ i.e., Enthalpy change on freezing of 18$$\,g$$ of water at $${0^ \circ }C = - 6.03\,kJ$$
∴ Enthalpy change on freezing 1000$$\,g$$ of water $$ = - \frac{{6.03}}{{18}} \times 1000 = - 335\,kJ$$
118.
The heat of atomization of methane and ethane are $$360\,kJ/mol$$ and $$620\,kJ/mol,$$ respectively. The longest wavelength of light capable of breaking the $$C – C$$ bond is ( Avogadro number $$ = 6.02 \times {10^{23}}mo{l^{ - 1}},h = 6.62 \times {10^{ - 34}}J\,\left. s \right):$$
119.
$$\left( {\Delta H - \Delta U} \right)$$ for the formation of carbon monoxide $$(CO)$$ from its elements at $$298 K$$ is
$$\left( {R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}} \right)$$
A
$$ - 2477.57\,J\,mo{l^{ - 1}}$$
B
$$2477.57\,J\,mo{l^{ - 1}}$$
C
$$ - 1238.78\,J\,mo{l^{ - 1}}$$
D
$$1238.78\,J\,mo{l^{ - 1}}$$
Answer :
$$1238.78\,J\,mo{l^{ - 1}}$$
For the reaction,
$$\eqalign{
& {C_{\left( s \right)}} + \frac{1}{2}{O_{2\left( g \right)}} \to CO \cr
& \Delta H = \Delta U + \Delta nRT\,\,or\,\,\Delta H - \Delta U = \Delta nRT \cr
& \Delta n = 1 - \frac{1}{2} = \frac{1}{2}; \cr
& \Delta H - \Delta U = \frac{1}{2} \times 8.314 \times 298 \cr
& = 1238.78\,J\,mo{l^{ - 1}} \cr} $$
120.
For a reaction, $$P + Q \to R + S.$$ The value of $$\Delta {H^ \circ }$$ is $$ - 30\,kJ\,mo{l^{ - 1}}$$ and $$\Delta S$$ is $$ - 100\,J\,{K^{ - 1}}\,mo{l^{ - 1}}.$$ At what temperature the reaction will be at equilibrium ?
A
$$27{\,^ \circ }C$$
B
$$52{\,^ \circ }C$$
C
$$30{\,^ \circ }C$$
D
$$45{\,^ \circ }C$$
Answer :
$$27{\,^ \circ }C$$
$$\eqalign{
& \Delta G = \Delta H - T\Delta S, \cr
& {\text{At equilibrium}},\Delta G = 0,\Delta H = T\Delta S \cr
& 30000 = T \times 100;T = 300\,K\,\,{\text{or}}\,\,27{\,^ \circ }C \cr} $$