Thermodynamics · Chemistry · NEET
MCQ (Single Correct Answer)
Choose the correct statement for the work done in the expansion and heat absorbed or released when 5 litres of an ideal gas at 10 atmospheric pressure isothermally expands into vacuum until volume is 15 litres :
For an endothermic reaction:
(A) $$\mathrm{q}_{\mathrm{p}}$$ is negative.
(B) $$\Delta_{\mathrm{r}} \mathrm{H}$$ is positive.
(C) $$\Delta_r \mathrm{H}$$ is negative.
(D) $$\mathrm{q}_{\mathrm{p}}$$ is positive.
Choose the correct answer from the options given below:
For the following reaction at $$300 \mathrm{~K}$$
$$\mathrm{A}_2(\mathrm{~g})+3 \mathrm{~B}_2(\mathrm{~g}) \rightarrow 2 \mathrm{AB}_3(\mathrm{~g})$$
the enthalpy change is $$+15 \mathrm{~kJ}$$, then the internal energy change is :
In which of the following processes entropy increases?
A. A liquid evaporates to vapour.
B. Temperature of a crystalline solid lowered from $$130 \mathrm{~K}$$ to $$0 \mathrm{~K}$$.
C. $$2 \mathrm{NaHCO}_{3(\mathrm{~s})} \rightarrow \mathrm{Na}_2 \mathrm{CO}_{3(\mathrm{~s})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$$
D. $$\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{Cl}_{(g)}$$
Choose the correct answer from the options given below:
Match List I with List II.
List I (Process) |
List II (Conditions) |
||
---|---|---|---|
A. | Isothermal process | I. | No heat exchange |
B. | Isochoric process | II. | Carried out at constant temperature |
C. | Isobaric process | III. | Carried out at constant volume |
D. | Adiabatic process | IV. | Carried out at constant pressure |
Choose the correct answer from the options given below:
The work done during reversible isothermal expansion of one mole of hydrogen gas at $$25^{\circ} \mathrm{C}$$ from pressure of 20 atmosphere to 10 atmosphere is (Given $$\mathrm{R}=2.0 \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$)
Consider the following reaction :-
$$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-483.64 \mathrm{~kJ} \text {. }$$
What is the enthalpy change for decomposition of one mole of water? (Choose the right option).
The equilibrium concentrations of the species in the reaction $$\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}$$ are $$2,3,10$$ and $$6 \mathrm{~mol}$$ $$\mathrm{L}^{-1}$$, respectively at $$300 \mathrm{~K} . \Delta \mathrm{G}^{0}$$ for the reaction is $$(\mathrm{R}=2 \mathrm{cal} / \mathrm{mol} ~\mathrm{K})$$
Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?
One mole of an ideal gas at 300 K is expanded isothermally from 1 L to 10 L volume. $$\Delta$$U for this process is :
(Use R = 8.314 J k$$-$$1 mol$$-$$1)
A vessel contains 3.2 g of dioxygen gas at STP (273.15 K and 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes one third of the original pressure. The volume of new vessel in L is : (Given : molar volume at STP is 22.4 L)
Which of the following p-V curve represents maximum work done?
Sucrose + H2O ⇌ Glucose + Fructose
If the equilibrium constant (Kc) is 2 $$ \times $$ 1013 at 800 K, the value of $$\Delta r{G^\Theta }$$ at the same temperature will be :
$$\Delta $$U = 2.1 kcal, $$\Delta $$S = 20 cal K$$-$$1 at 300 K
Hence, G is
(i) C(graphite) + O2(g) $$ \to $$ CO2(g); $$\Delta $$rHo = x kJ mol$$-$$1
(ii) C(graphite) + $${1 \over 2}$$O2(g) $$ \to $$ CO(g); $$\Delta $$rHo = y kJ mol$$-$$1
(iii) CO(g) + $${1 \over 2}$$O2(g) $$ \to $$ CO2(g); $$\Delta $$rHo = z kJ mol$$-$$1
Based on the above equations, find out which of the relationship given below is correct.
H (kJ/mol) | |
---|---|
1/2A $$ \to $$ B | +150 |
3B $$ \to $$ 2C + D | -125 |
E + A $$ \to $$ 2D | +350 |
For B + D $$ \to $$ E + 2C, $$\Delta $$H will be
4H(g) $$ \to $$ 2H2(g) is $$-$$869.6 kJ
The dissociation energy of H $$-$$ H bond is
List I | List II | ||
---|---|---|---|
Equations | Type of processes | ||
A. | Kp > Q | (i) | Non- spontaneous |
B. | $$\Delta $$Go < RT ln Q | (ii) | Equilibrium |
C. | Kp = Q | (iii) | Spontaneous and endothermic |
D. | T > $${{\Delta H} \over {\Delta S}}$$ | (iv) | Spontaneous |
Fe2O3(s) + 3CO(g) $$ \to $$ 2Fe(s) + 3CO2(g); $$\Delta $$H = $$-$$ 26.8 kJ
FeO(s) + CO(g) $$ \to $$ Fe(s) + CO2(g); $$\Delta $$H = $$-$$ 16.5 kJ
The value of $$\Delta $$H for the following reaction
Fe2O3(s) + CO(g) $$ \to $$ 2FeO(s) + CO2(g) is
1/2X2 + 3/2Y2 $$\rightleftharpoons$$ XY3, $$\Delta $$H = $$-$$ 30 kJ,
to be at equilibrium, the temperature should be
H $$-$$ H bond energy : 431.37 kJ mol$$-$$1
C $$=$$ C bond energy : 606.10 kJ mol$$-$$1
C $$-$$ C bond energy : 336.49 kJ mol$$-$$1
C $$-$$ H bond energy : 410.50 kJ mol$$-$$1
Enthalpy for the reaction,
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will be
C(graphite) + CO2(g) $$ \to $$ 2CO(g)
are 170 kJ and 170 J K$$-$$1, respectively. This reaction will be spontaneous at
PCl5(g) $$\rightleftharpoons$$ PCl3(g) + Cl2(g)
which of the following conditions are correct ?
(I) q + w (II) q
(III) w (IV) H $$-$$ TS
(i) H+(aq) + OH$$-$$(aq) = H2O(l), $$\Delta $$H = $$-$$ X1 kJ mol$$-$$1
(ii) H2(g) + 1/2O2(g) = H2O(l), $$\Delta $$H = $$-$$ X2 kJ mol$$-$$1
(iii) CO2(g) + H2(g) = CO(g) + H2O(l), $$\Delta $$H = $$-$$ X3 kJ mol$$-$$1
(iv) C2H2(g) + 5/2O2(g) = 2CO2(g) + H2O(l), $$\Delta $$H = +X4 kJ mol$$-$$1
Enthalpy of formation of H2O(l) is
Br2(l) + Cl2(g) $$ \to $$ 2BrCl(g)
are 30 kJ mol$$-$$1 and 105 J K$$-$$1 mol$$-$$1 respectively.
The temperature at which the reaction will be in equilibrium is
Mg(O)(s) + 2HCl(aq) $$ \to $$ MgCl2(aq) + H2O(l) will be
H2(g) + Br2(g) $$ \to $$ 2HBr(g) is
(i) Pure solvent $$ \to $$ separated solvent molecules, $$\Delta $$H1
(ii) Pure solute $$ \to $$ separated solute molecules, $$\Delta $$H2
(iii) Separated solvent and solute molecules $$ \to $$ solution, $$\Delta $$H3
Solution so formed will be ideal if
C3H8(g) + 5O2(g) $$ \to $$ 3CO2(g) + 4H2O(l)
at constant temperature, $$\Delta $$H $$-$$ $$\Delta $$E is
2H2O2(l) $$ \to $$ 2H2O(l) + O2(g)
if heat of formation of H2O2(l) and H2O(l) are $$-$$188 and - 286 kJ/mol respectively, is
SnO2 $$ \to $$ SnO; $$\Delta $$G298 > 0
Most probable oxidation state of Pb and Sn will be
C2H5OH(l) + 3O2(g) $$ \to $$ 2CO2(g) + 3H2O(l)
which one is true
2Zn + S2 $$ \to $$ 2ZnS; $$\Delta $$Go = $$-$$ 293 J
S2 + 2O2 $$ \to $$ 2SO2; $$\Delta $$Go = $$-$$408 J
$$\Delta $$Go for the following reaction
2ZnS + 3O2 $$ \to $$ 2ZnO + 2SO2 is