CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 9 - MCQExams.com

Which of the following parameters cannot be estimated by using the Born-Haber cycle?
  • Hydration energy of the ion
  • Electron gain enthalpy
  • Lattice energy
  • Electronegativity
Born-Haber cycle can be used to estimate :
  • Lattice energy of ionic crystals
  • Electron gain enthalpy
  • Electronegativity
  • Both (A) and (B)
Which one of the following gases possesses the largest internal energy?
  • 2 moles of helium occupying $$1m^{3}$$ at 300 K
  • 56 kg of nitrogen at 107$$Nm^{-2}$$ and 300 K
  • 8 grams of oxygen at 8 atm and 300 K
  • $$6\times 10^{26}$$ molecules of argon occupying $$4m^{3}$$ at 900 K
Which of the following relationship is not correct for the relation between $$\Delta H$$ and $$\Delta  U$$?
  • When $$\Delta n_{g} = 0$$ then $$\Delta H = \Delta U$$
  • When $$\Delta n_{g} > 0$$ then $$\Delta H > \Delta U$$
  • When $$\Delta n_{g} < 0$$ then $$\Delta H < \Delta U$$
  • When $$\Delta n_{g} = 0$$ then $$\Delta H > \Delta U$$
For the electrochemical cell, $$ \mathrm{M} | \mathrm{M}^{+} \| \mathrm{X}^{-} \mid \mathrm{X}, $$
$$ E_{M}^{o}+/ M=0 \cdot 44 V $$ and $$ E_{x / x^-}^{\circ}=0 \cdot 33 V $$From these data one can deduce that
  • $$ M+X \rightarrow M^{+}+X^{-} $$ is the spontaneous reaction
  • $$ M^{+}+X^{-} \rightarrow M+X $$ is the spontaneous reaction
  • $$ E_{c e l l}=0 \cdot 77 V $$
  • $$ \mathrm{E}_{\mathrm{cell}}=-0.77 \mathrm{V} $$
Hess's law is applicable for the determination of heat of
  • transition
  • formation
  • reaction
  • all of these
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below $$\displaystyle \frac{1}{2}\mathrm{C}1_{2}(\mathrm{g})$$ to $$\mathrm{C}1^{-}(\mathrm{a}\mathrm{q})$$
(using the data, $$\Delta_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}}\mathrm{H}_{\mathrm{C}1_{2}}^{\Theta}=240\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1},\ \Delta_{\mathrm{e}\mathrm{g}}\mathrm{H}_{\mathrm{C}1}^{\Theta}=-349\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1},\ \Delta_{\mathrm{h}\mathrm{y}\mathrm{d}}\mathrm{H}_{\mathrm{C}1^{-}}=-381\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$) will be 
  • $$-850\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$
  • $$+120\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$
  • $$+152\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$
  • $$-610\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$

Standard enthalpy and standard entropy for the oxidation of $$NH_3$$ at $$298$$ K are $$-382.64$$ $$\text{kJ}\ \text{mol}^{-1}$$ and $$-145.6 $$ $$\text{J}\ {\text{mol}}^{-1}\ \text K^{-1}$$ respectively. Standard free energy change (in $$\text{kJ mol}^{-1} $$) for the same reaction at $$298$$ K is:
  • $$-221.1$$
  • $$-339.3$$
  • $$-439.3$$
  • $$-523.2$$
For the equilibrium : $$CaC{O}_{3}\left(s\right) \rightleftharpoons CaO\left(s\right) + C{O}_{2}\left(g\right)$$, the pressure of $$C{O}_{2}$$ is 1 atm at equilibrium in a container of $$V$$ litre, at temperature $$T$$. Which one is not correct?
  • $$T = \displaystyle\frac{\Delta H}{\Delta S}$$
  • $$\Delta G = 1$$
  • $$\Delta G = 0$$
  • $$T = \displaystyle\frac{V}{nR}$$

Assertion (A): At equilibrium $$\Delta G$$ becomes zero.

Reason (R):  At equilibrium the two tendencies $$\Delta H$$and $$T\Delta S$$ become equal.
  • Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
  • Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
  • Assertion is true but Reason is false.
  • Assertion is false and Reason is true.

The standard Gibbs energy change for the formation of propane $$(C_3H_{8(g)})$$ at 298 K is

$$[\Delta _{f}H^{\theta }$$ for propane $$=-103.85$$ kJ mole$$^{-1},$$ $$S^{\theta }_{m}\;C_{3}H_{8(g)}=270.2$$ JK$$^{-1}$$mol$$^{-1}$$, $$S^{\theta }_{m}\;H_{2(g)}=130.68$$ JK$$^{-1}$$mol$$^{-1}$$, $$S^{\theta}_{m}C_{(graphite)}=5.742$$ JK$$^{-1}$$mol$$^{-1}]$$
  • $$-23.4$$ kJ
  • $$-44.4$$ kJ
  • $$-54.4$$ kJ
  • $$-104.5$$ kJ

Match the list -I with list - II.

List-I                                                                    List - II

A) Hess law is not applicable for                          1) is not a state function

B) All combustion reactions are                           2) Heat of dilution

C) Work                                                                  3) Exothermic

D) Difference between two integral                     4) Nuclear reaction

    heats of solutions
  • A-1, B-2, C-3, D-4
  • A-4, B-3, C-1, D-2
  • A-3, B-2, C-4, D-1
  • A-2, B-1, C-4, D-3

Standard entropy of $$X_{2}$$, $$Y_{2}$$ and $$XY_{3}$$ are 60, 40 and 50 $$JK^{-1}mol^{-1}$$ respectively. For the reaction: $$\dfrac{1}{2}X_{2}+\dfrac{3}{2}Y_{2}\rightarrow XY_{3}, \Delta H=-30\: kJ$$ to be at equilibrium, the temperature will be :
  • 1250 K
  • 500 K
  • 1000 K
  • 750 K

The value $$\Delta $$S for the process, $$H_{2}O(s) \rightarrow H_{2}O(l)$$ at 1 atm pressure and 260K is greater than zero. The value of $$\Delta G$$ will be :
  • $$>0$$
  • $$<0$$
  • $$=0$$
  • lies between $$-1$$ and $$0$$

A coffee cup calorimeter initially contains 125g of water at a temperature of $$24.2^{0}C$$. After 10.5gm solute is added to the water at the same temperature, the final temperature becomes $$21.1^{0}C$$. The heat of solution is:
  • 85 J/g
  • 110 J/g
  • 270 J/g
  • 167 J/g

The heat of combustion for $$C$$, $$H_{2}$$ and $$CH_{4}$$ are $$-349$$ kJ/mol, $$-241.8$$ kJ/mol and $$-906.7$$ kJ/mol respectively. The heat of formation in kJ/mol of $$CH_{4}$$ is :
  • $$174.1$$
  • $$274.1$$
  • $$374.1$$
  • $$74.1$$
Select the incorrect statement(s) :
  • Lattice energy $$\propto \frac{1}{r^2}$$ {where r is the interionic distance}.
  • Lattice energy $$\propto$$ $$q_1q_2$$. {where $$q_1$$ and $$q_2$$ are the charges of co-ions}.
  • Ionic mobility in aqueous state  $$\propto \frac{1}{radius\ of\ ions\ in\ gaseous\ state}$$
  • Heat of formation of a compound depends on the number of steps involved in its formation reaction.
The change in free energy accompanied by the isothermal reversible expansion of 1 mole of an ideal gas when it doubles its volume is $$\Delta G_{1}$$. The change in free energy accompanied by a sudden isothermal irreversible doubling volume of 1 mole of the same gas is $$\Delta G_{2}$$. The ratio of $$\Delta G_{1}$$ and $$\Delta G_{2}$$ is :
  • $$1$$
  • $$\dfrac{1}{2}$$
  • $$2$$
  • $$\dfrac{3}{2}$$
Consider the following processes.
$$\Delta H (kJ/mol)$$
$$1/2A\rightarrow B$$+150
$$3B\rightarrow 2C+D$$-125
$$E+A\rightarrow 2D$$+350
For $$B+D\rightarrow E+2C, \Delta H$$ will be:
  • 325 kJ/mol
  • 525 kJ/mol
  • -175 kJ.mol
  • -325 kJ/mol
In conversion of lime-stone to lime, $$CaCO_{3}(s)\rightarrow CaO(s) + CO_{2}(g)$$, the values of $$\Delta^0H$$ and $$\Delta S^0 $$ are $$+179.1\ kJ mol^{-1}$$ and $$160.2\ J/Kmol$$ respectively at $$298\ K$$ and $$1$$ bar. Assuming that $$\Delta H^o$$ and $$\Delta S^o$$ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is:
  • $$1008\ K$$
  • $$1200\ K$$
  • $$845\ K$$
  • $$1118\ K$$
The values of $$\Delta H$$ and $$\Delta S$$ for the reaction, C(graphite) + $$CO_2(g)\rightarrow 2CO(g)$$ are 170 kJ and $$170\; JK^{-1}$$ respectively. This reaction will be spontaneous at:
  • 510 K
  • 710 K
  • 910 K
  • 1010 K
A reaction occurs spontaneously if:
  • T$$\Delta $$S > $$\Delta $$H and $$\Delta $$H is +ve and $$\Delta $$S is -ve
  • T$$\Delta $$S = $$\Delta $$H and both $$\Delta $$H and $$\Delta $$S are +ve
  • T$$\Delta $$S < $$\Delta $$H and both $$\Delta $$H and $$\Delta $$S are +ve
  • T$$\Delta $$S > $$\Delta $$H and both $$\Delta $$H and $$\Delta $$S are +ve
A 1000 gm sample of water is reacted with an equimolar amount of CaO (both at same initial temperature of $$25^{\circ}C$$). Assuming the container to be adiabatic, the final temperature of the product is approx :

Given : $$CaO + H_2O\rightarrow Ca(OH)_2,\ \Delta H=-65.2 KJ/mol $$ 
Specific heat of $$Ca(OH)_2 = 1.2 \: J/gm^{\circ}C$$
  • $$735^{\circ}C$$
  • $$760^{\circ}C$$
  • $$746^{\circ}C$$
  • $$789^{\circ}C$$
If $$\Delta H{_{f}}^{0}$$ for $$Ag^{+} (\infty$$  diluted), $$NO{_{3}}^{-} (\infty$$ diluted), $$Cl^{-} (\infty$$ diluted) and $$AgCl_{(S)}$$ are $$-105.579, -207.36, -167.159$$ and $$-127.068$$ respectively. Calculate the enthalpy change for the reaction $$AgNO_{3(aq)}+HCl_{(aq)}\rightarrow AgCl_{(S)}+HNO_{3} (aq)$$.
  • 21.471 kJ/mol
  • 145.67 kJ/mol
  • 65.488 kJ/mol
  • None
$$C(s) + O_2(g)  \rightarrow  CO_2(g),  \quad \Delta{H} = + 94.0  K cal.$$

$$CO(g) +\frac{1}{2} O_2(g)   \rightarrow CO_2(g)    ,\quad \Delta H = -67.7  K  cal.$$

From the above reactions find how much heat (Kcal mole$$^{-1}$$) would be produced in the following reaction:

$$C(s) + \frac{1}{2} O_2(g)  \rightarrow CO(g)$$
  • 20.6
  • 26.3
  • 44.2
  • 161.7
Given that
$$Zn + \frac{1}{2} O_2  \rightarrow ZnO\quad\Delta{H}= + 84000  cal$$      ........... 1
$$Hg+\frac{1}{2} O_2  \rightarrow HgO\quad \Delta{H} =+ 21700  cal$$       ...........  2

The heat of reaction $$(\Delta H)$$ for,

$$Zn + HgO  \rightarrow ZnO + Hg$$ is
  • 105700 cal
  • 62300 cal
  • -105700 cal
  • -62300 cal
Specific heat of substance at m.pt. (or b.pt. or for isothermal process) and specific heat of a substance during adiabatic change respectively are
  • 0, 0
  • 0, $$\infty$$
  • $$\infty$$, 0
  • $$\infty, \infty$$
The heat of reaction for $$A + \frac{1}{2} O_2  \rightarrow AO$$ is $$-50$$ kcal/mol and $$AO + \frac{1}{2} O_2  \rightarrow AO_2$$ is $$100$$ kcal/mol.
The heat of reaction (in kcal/mol) for $$A + O_2  \rightarrow AO_2$$ will be:
  • $$-50$$
  • $$+50$$
  • $$+100$$
  • $$+150$$
For an ideal gas, enthalpy becomes: 
  • $$\Delta H = \Delta U - RT$$
  • $$\Delta H = -\Delta U - RT$$
  • $$\Delta H = \Delta U + RT$$
  • $$\Delta H = -\Delta U + RT$$
Consider the following data: $$\Delta_fH^o(N_2H_4, l)=50kJ/mol, \Delta_fH^o(NH_3, g)=-46 kJ/mol$$,
$$B.E. (N-H)=393 kJ/mol$$ and $$B.E. (H-H)=436 kJ/mol$$,
$$\Delta_{vap}H(N_2H_4, l)=18 kJ/mol$$
Calculate the $$N-N$$ bond energy in kJ/mol for $$N_2H_4$$.
  • $$190 \ kJ/mol$$
  • $$-190 \ kJ/mol$$
  • $$95 \ kJ/mol$$
  • $$-95 \ kJ/mol$$
Following are the thermochemical reactions:
$$C$$ (graphite) $$+ \dfrac{1}{2}O_2  \rightarrow CO \ ;  \Delta H = - 110.5$$ kJ/mol
$$CO + \dfrac{1}{2} O_2  \rightarrow CO_2 \ ;  \Delta H = - 283.2$$ kJ/mol
The heat of reaction (in kJ/mol) for the following reaction is:
$$C (\text{graphite})  + O_2  \rightarrow CO_2$$
  • $$+393.7$$
  • $$-393.7$$
  • $$-172.7$$
  • $$+172.7$$
Find the heat change in the reaction
$$NH_3(g) + HCl (g)   \rightarrow NH_4 Cl(s)$$ from the following data
$$NH_3 (g) + aq  \rightarrow NH_3 (aq),$$                    $$\Delta H = - 8.4  K.  Cal.$$
$$HCl(g) + aq  \rightarrow HCl(aq),$$                     $$\Delta H = -17.3 K.  Cal.$$
$$NH_3(aq) + HCl(aq) \rightarrow NH_4 Cl(aq),  \Delta H = - 12.5  K. Cals.$$
$$NH_4 Cl(s) + aq \rightarrow NH_4 Cl(aq),                \Delta H = + 3.9  K.  Cal.$$
  • -42.1
  • -23.3
  • +34.3
  • +42.1
For the reaction : $$X_2O_4(l)\rightarrow 2XO_2(g)$$. $$\Delta U = 2.1\,kcal,\  \Delta S = 20\,cal\,K^{ -1 }\,at\ 300\,K$$. 
Hence $$\Delta G$$ is:
  • $$-7\,kcal$$
  • $$-4.7\,kcal$$
  • $$+2.7\,kcal$$
  • $$-2.7\,kcal$$
200 ml of 1M HCl is mixed with 400 ml of 0.5M NaOH. The temperature rise in the calorimeter was found to be 4.0oC. Water equivalent of calorimeter is 25g and the specific heat of the solution is 1 cal/mL/degree. If the theoritical heat of neutralization of a strong acid and strong base is −13.5 kcal, then the percentage error  in this experiment while calculating the heat of neutralization is
  • 7.4
  • 3
  • 5.6
  • 4.5
For the hypothetical reaction: 
   $${ A }_{ 2 }(g)+{ B }_{ 2 }(g)\rightleftharpoons 2AB(g)$$

$${ \Delta  }_{ r }{ G }^{ o }$$ and $${ \Delta  }_{ r }{ S }^{ o }$$ is $$20 kJ/mol$$ and $$-20{ JK }^{ -1 }{ mol }^{ -1 }$$ respectively at $$200K$$. If $${ \Delta  }_{ r }{ C  }_{ p }$$ is $$20{ JK }^{ -1 }{ mol }^{ -1 }$$ then $${ \Delta  }_{ r }{ H }^{ o }$$ at $$400K$$ is :
  • $$20\ kJ/mol$$
  • $$7.98\ kJ/mol$$
  • $$28\ kJ/mol$$
  • none of these
Using the listed information calculate $${ \Delta  }_{ r }{ G }^{ o }$$ (in kJ/mol) at $${ 27 }^{ o }C$$.
$${ Co }_{ 3 }{ O }_{ 4 }(s)+4CO(g)\longrightarrow 3Co(s)+4{ CO }_{ 2 }(g)$$

Given : At $$300K$$ , $$\Delta { { H }^{ o } }_{ f }(kJ/mol) $$ are $$-891,\ -110.5,\ 0.0$$ and $$-393.5$$ respectively.
           $${ S }^{ o }(J/K.mol)$$ are $$102.5,\ 197.7, \ 30.0$$ and $$213.7$$ respectively.
  • $$214.8$$
  • $$-195.0$$
  • $$-200.3$$
  • $$-256.45$$
0.16 g of methane was subjected to combustion at $$27^{\circ}C$$ in bomb calorimeter. The temperature of calorimeter system (including water)  was found to rise by $$0.5^{\circ}C$$. If the heat of combustion of methane at constant volume and constant pressure is $$x$$ kJ/mole, find the value of $$x$$. The thermal capacity of the calorimeter system is $$17.7\ kJ\ K^{-1},\ R=8.314\ JK^{-1}\ mol^{-1}$$.
  • $$-890$$
  • $$890$$
  • $$450$$
  • $$-450$$
Determine $${ \Delta  }{ U }^{ o }$$ at $$300K$$ for the following reaction using the listed enthalpies of reaction:

$$4CO(g)+8{ H }_{ 2 }(g)\longrightarrow 3{ CH }_{ 4 }(g)+{ CO }_{ 2 }(g)+2{ H }_{ 2 }O(l)$$

$$C_{(graphite)}+1/2{ O }_{ 2 }(g)\longrightarrow CO(g);\quad \Delta { { H }_{ 1 } }^{ o }=-110.5kJ$$

$$CO(g)+1/2{ O }_{ 2 }(g)\longrightarrow { CO }_{ 2 }(g);\quad \Delta { { H }_{ 2 } }^{ o }=-282.9kJ$$

$${ H }_{ 2 }(g)+1/2{ O }_{ 2 }(g)\longrightarrow { H }_{ 2 }O(l);\quad \Delta { { H }_{ 3 } }^{ o }=-285.8kJ$$

$$C_{(graphite)}+2{ H }_{ 2 }(g)\longrightarrow { CH }_{ 4 }(g);\quad \Delta { { H }_{ 4 } }^{ o }=-74.8kJ$$
  • $$653.5\ kJ$$
  • $$-686.2\ kJ$$
  • $$-747.4\ kJ$$
  • None of these
The enthalpy changes of the following reactions at $$27^{\circ}C$$ are 
$$\displaystyle \:Na(s)+\frac{1}{2}Cl_{2}(g)\rightarrow NaCl(s)$$$$\displaystyle \:\Delta _{r}H= -411 kJ/mol$$
$$\displaystyle \:H_{2}(g)+S(s)+2O_{2}(g)\rightarrow H_{2}SO_{4}(l)$$$$\displaystyle \:\Delta _{r}H= -811 kJ/mol$$
$$\displaystyle \:2Na(s)+S(s)+2O_{2}(g)\rightarrow Na_{2}SO_{4}(s)$$$$\displaystyle \:\Delta _{r}H= -1382 kJ/mol$$
$$\displaystyle \:\frac{1}{2}H_{2}(g)+\frac{1}{2}Cl
_{2}(g) \rightarrow HCl(g)$$		$$\displaystyle \:\Delta _{r}H= -92 kJ/mol;$$
from these data, the heat change of reaction at constant volume (in kJ/mol) at $$27^{\circ}C$$ for the process $$\displaystyle \:2NaCl(s)+H_{2}SO_{4}(l)\rightarrow Na_{2}SO_{4}(s)+2HCl(g)$$
  • 67
  • 62.02
  • 71.98
  • None
Combustion of surose is used by aerobic organisms for providing energy for the life sustaining processes. If all the capturing of energy from the reaction is done through electrical process (non P-V work) then calculate maximum available energy which can be captured by combustion of $$34.2gm$$ of sucrose
Given : $$\Delta {{H}_{combustion}}(sucrose)=-6000\ kJ. {mol}^{-1}$$
             $$\Delta {{S}_{combustion}}=180J/K.mol$$ and body temperature is $$300K$$
  • $$600kJ$$
  • $$594.6kJ$$
  • $$5.4kJ$$
  • $$605.4kJ$$
Find the forward$$\displaystyle \:\Delta _{r}U^{\circ}$$ at 300K for the reaction  $$4HCl(g)+O_{2}(g)\rightarrow 2Cl_{2}(g)+2H_{2}O(g)$$ Assume all gases are ideal. Given $$\displaystyle \:H_{2}(g)+Cl_{2}(g)\rightarrow 2HCl(g)$$ $$\Delta _{r}H_{300}^{^{\circ}}= -184.5 kJ/mol$$ 
$$\displaystyle \:2H_{2}(g)+O_{2}(g)\rightarrow 2H_2O(g)$$ $$\Delta _{r}H_{300}^{^{\circ}}= -483 kJ/mol$$ (Use R= 8.3 J/mole)
  • 111.5 kJ/mole
  • -109.01 kJ/mole
  • -111.5 kJ/mole
  • None
The reaction that proceeds in the forward direction is:
  • $$SnC{l}_{4} + H{g}_{2}C{l}_{2} \longrightarrow SnC{l}_{2} + 2HgC{l}_{2}$$
  • $$N{H}_{4}Cl + NaOH \longrightarrow {H}_{2}O + N{H}_{3} + NaCl$$
  • $${Mn}^{2+} + 2{H}_{2}O + C{l}_{2} \longrightarrow Mn{O}_{2} + 4{H}^{+} + 2C{l}^{-}$$
  • $${S}_{4}{O}_{6}^{2-} + 2{I}^{-} \longrightarrow 2{S}_{2}{O}_{3}^{2-} + {I}_{2}$$
In the Born-Haber cycle for the formation of solid common salt ($$Nacl$$), the largest contribution comes from
  • the low ionization potential of $$Na$$
  • the high electron affinity of $$Cl$$
  • the low $$\Delta H_{vap}$$ of $$Na(s)$$
  • the lattice energy
Use the following data to calculate second electron ainity of oxygen, i.e., for the process
$$O^{-}(g) + e^{-}(g)  \rightarrow O^{2-}(g)$$
Is the $$O^{2-}$$ ion stable in the gas phase?.Why is it stable in solid MgO?
Heat of sublimation of $$Mg(s) = + 147.7 kJ mol^{-1}$$
Ionisation energy of Mg(g) to form
$$Mg^{2+}(g) = + 2189.0 kJ mol^{-1}$$
Bond dissociation energy for $$O_2 = + 498.4 kJmol^{-1}$$
First electron affinity of $$O(g) = - 141.0 kJ mol^{-1}$$
Heat formation of $$MgO(s) = -601.7 kJ mol^{-1}$$
Lattice energy of $$MgO = -3791.0 kJ mol^{-1}$$
  • 601.7
  • 744.4
  • 1346.1
  • 147.7

The energy change for the alternating reaction that yields chlorine sodium $$(Cl^{+}Na^{-})$$ will be:

$$2Na(s)\, +\, Cl_2(g)\,\rightarrow\, 2Cl^{+}Na^{-}(s)$$

Given that:

Lattice energy of $$NaCl\,=\,-787\, kJ\,mol^{-1}$$

Electron affinity of $$Na\,=\,-52.9\, kJ\, mol^{-1}$$

Ionisation energy of $$Cl\, =\, +\,1251\, kJ\, mol^{-1}$$

BE of $$Cl_2\,=\,244\, kJ\, mol^{-1}$$

Heat of sublimation of $$Na(s)\, =\,107.3\, kJ\, mol^{-1}$$

$$\Delta H_f(NaCl)\, =\,-411\, kJ\, mol^{-1}$$.
  • +640 kJ
  • +1280 kJ
  • -410 kJ
  • +410 kJ
Select correct statement.
  • Both lattice energy and hydration energies decrease with ionic size.
  • Lattice energy can be calculated using Born-Haber cycle.
  • If the anion is larger compared to the cation, the lattice energy will remain almost constant within a particular group.
  • All the above are correct statements.

The lattice energy of NaCl(s) using the following data will be:

heat of sublimation of $$Na(s)\,=\,S$$

$$(IE)_1$$ of $$Na\,(g)\,=\,I$$

bond dissociation energy of $$Cl_2\,(g)\,=\,D$$

electron affinity of $$Cl\,(g)\,=\,-E$$

heat of formation of $$NaCl(s)\,=\,-Q$$
  • Lattice energy $$-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$$
  • Lattice energy $$-U\, =\, S\, -\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$$
  • Lattice energy $$-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, +\, E\, -\,Q$$
  • Lattice energy $$-U\, =\, S\, -\, I\, -\,\displaystyle \frac{D}{2}\, +\, E\, +\,Q$$
An ideal monoatomic gas undergoes a process in which its internal energy U and density $$\rho$$ vary as $$U\rho\, =\, constant.$$ The ratio of change in internal energy and the work done by the gas is
  • $$\displaystyle \frac{3}{2}$$
  • $$\displaystyle \frac{2}{3}$$
  • $$\displaystyle \frac{1}{3}$$
  • $$\displaystyle \frac{3}{5}$$
The value of $$\Delta U$$ for the reaction $$2A\left(g\right) + B\left(g\right) \rightleftharpoons {A}_{2}B\left(g\right)$$ for which $${K}_{p} = 1.0\times {10}^{-10}  {atm}^{-2}$$ and $$\Delta S = 5  J{K}^{-1}$$ and $$T = 300  K$$, is :
  • 53.93 kJ
  • 63.93 kJ
  • 56.24 kJ
  • 68.24 kJ
1.0 g magnesium atoms in vapour phase absorbs 50.0 kJ of energy to convert all Mg into Mg ions. The energy absorbed is needed for the following changes :

           $$Mg(g)\longrightarrow Mg^+ (g)+ e; \:  \Delta H = 740 \: kJ \: mol^{-1}$$
       $$Mg^+(g)\longrightarrow Mg^{2+} (g)+ e; \:  \Delta H = 1450 \: kJ \: mol^{-1}$$

Find out the % of $$Mg^+$$ and $$Mg^{2+}$$ in the final mixture.
  • % $$Mg^+ = 68.28$$% , % $$of Mg^{2+} = 31.72$$%
  • % $$Mg^+ =58.28$$% , % $$of Mg^{2+} = 41.72$$%
  • % $$Mg^+ = 78.28$$% , % $$of Mg^{2+} = 21.72$$%
  • None of these
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