4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 76 InterviewSolution

3751.	Which one is the correct unit for entropy? (a) KJ mol (b) JK-1 mol (c) JK-1 mol-1 (d) KJ mol-1
Answer» (c) JK^-1 mol^-1

Discussion

3752.	30.4 KJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK-1 mol-1 Calculate the melting point of sodium chloride.
Answer» Heat required for 1 mole of NaCl for melting (q) = 30.4 K J = 30.4 x 1000 J ∆S – entropy change = 28.4 J K^-1 mol^-1 Melting point = T_m = ? ∆S = q = T_m ∴ T_m= q / ΔS T_m= (30.4 x 1000) / 28.4 = 10704 K Melting point of NaCl = 1070.4 K.

3752.

30.4 KJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK-1 mol-1 Calculate the melting point of sodium chloride.

Answer»

Heat required for 1 mole of NaCl for melting (q) = 30.4 K J

= 30.4 x 1000 J

∆S – entropy change = 28.4 J K^-1 mol^-1

Melting point = T_m = ?

∆S = q = T_m

∴ T_m= q / ΔS

T_m= (30.4 x 1000) / 28.4 = 10704 K

Melting point of NaCl = 1070.4 K.

Discussion

3753.	Calculate the equilibrium constant for the reaction given below at `400K`, if `DeltaH^(Theta) = 77.2 kJ mol^(-1)` and `DeltaS^(Theta) = 122 J K^(-1) mol^(-1)` `PCI_(5)(g) rarr PCI_(3)(g) +CI_(2)(g)`
Answer» `DeltaH^(Theta) = 77.2 mol^(-1)` `DeltaS^(Theta) = 122 J K^(-1) mol^(-1)` `T = 400K` `:. DeltaG^(Theta) - T DeltaS^(Theta)` `DeltaG^(Theta) = 77200 - 400 xx122 = 28400 J` Also, we have `-DeltaG^(Theta) = 2.303 RT log_(10)K_(c)` where `K_(c)` is equilibrium constant `:. -28400 = 2.303 xx 8.314 xx 400 log_(10)K_(c)` `:. K_(c) = 1.958 xx 10^(-4)`

Discussion

3754.	Calculate the equilibrium constnat for the reaction given below at `400K`, if `DeltaH^(Theta) = 77.2 kJ "mole"^(-1)` and `DeltaS^(Theta) = 122 J K^(-1) "mole"^(-1)`. `PCI_(5)(g) rarr PCI_(3)(g) +CI_(2)(g)`
Answer» `DeltaH^(Theta) = 77.2 kJ mol^(-1), DeltaS^(Theta) = 122 J K^(-1) mol^(-1)` `T = 400K` `:. DeltaG^(Theta) = DeltaH^(Theta) - T DeltaG^(Theta)` `DeltaG^(Theta) = 77200 - 400 xx122 = 28400J` Also we have, `-DeltaG^(Theta) = 2.303 RT log K_(c)` where `K` is equilibrium constant `:. -28400 = 2.303 xx 8.314 xx 400 logK_(c)` `:. K_(c) = 1.958 xx 10^(-4)`

Discussion

3755.	The temperature in K at which `DeltaG=0` for a given reaction with `DeltaH=-20.5 kJ mol^(-1)` and `DeltaS=-50.0 JK^(-1) mol^(-1)` isA. `-410`B. `410`C. `2.44`D. `-2.44`
Answer» Correct Answer - B `DeltaG=DeltaH - T Delta S`

Discussion

3756.	Calculate the value of `DeltaG` at `700K` for the reaction, `nX to mB`. Given that value of `DeltaH=-113kJ mol^(-1)` and `DeltaS=-145JK^(-1)mol^(-1).`
Answer» `DeltaG=DeltaH-TDeltaS` `=-113xx10^(3)-700xx(-145) J mol^(-1)` `DeltaG=-113000+101500` `=-11500 J mol^(-1)` `DeltaG=-11.50 k J mol^(-1)`

Discussion

3757.	If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride vapor, thenA. `Delta H = Delta U`B. `Delta H gt Delta U`C. `Delta H lt Delta U`D. there is no relationship
Answer» Correct Answer - C Themochemical equation is `NH_(3) (g) + HCl (g) rarr NH_(4) Cl_(g)` Accroding to thermodynamics, `Delta H = Delta U + Delta n_(g) RT` where `Delta n_(g) = sum n_(P) (g) - sum n_(R ) (g)` `= (1) - (1 + 1)` `= -1` Since `Delta n_(g)` is negative `(lt 0)` we conclude that `Delta H lt Delta U`

Discussion

3758.	In thermodynamics, a process is called reversible whenA. the system changes into the surrounding spontaneouslyB. the surrounding are always in equilibrium with the systemC. there is no boundary between and surroundingD. surroundings and system change into each other
Answer» Correct Answer - B During a reversible process in themodynamics, the system and its surrounding are more or less in equilibrium because th process is carried out infinitesimally very slowly.

Discussion

3759.	If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporization of 1 mol of water at 1 bar and `100^(@)`C is 41 kJ `mol^(-1)`. Calculate the internal energy, when 1 mol of water is vapourised at one bar pressure and `100^(@)`C.
Answer» The change `H_(2)O(I) rarr H_(2)O(g)` `Delta H = Delta U + n_(g)RT` `or Delta U = Delta H - Delta n_(g)RT` = 41 kJ `mol^(-1) - 1 xx "8.314 J"" " mol^(-1)K^(-1)xx 373 K` = 41 kJ `mol^(-1)` - 3101 J `mol^(-1)` = 41 kJ`mol^(-1)` - 3.101 kj `mol^(1)` = 37.9 kJ `mol^(-1)`

Discussion

3760.	Which among the following properties is/are extenstive properties?A. MoleB. Heat capacityC. Molar enthalpyD. Entropy
Answer» Correct Answer - a,b,d

Discussion

3761.	A liquid has a vapour pressure of 40mm Hg at `19.0^(@)C` and a normal boiling point of `78.3^(@)C` What is its enthalpy of vaporization in KJ .`mol^(-1)`?A. 42.4B. 18.4C. `5.10`D. 1.45
Answer» Correct Answer - a

Discussion

3762.	A vessel of volume `V = 20 1` contains a mixture of hydrogen and helium at a temperature `t = 20 ^@C` and pressure `p = 2.0 atm`. The mass of the mixture is equal is equal to `m = 5.0 g`. Find the ratio of the mass of hydrogen to that of helium in the given mixture.
Answer» Let the mixture contain `v_1` and `v_2` moles of `H_2` and `H_e` respectively. If molecular weights of `H_2` and `H_e` are `M_1` and `M_2`, then respective masses in the mixture are equal to `m_1 = v_1 M_1 and m_2 = v_2 M_2` Therefore, for the total mass of the mixture we get, `m = m_1 + m_2` or `m = v_1 M_1 + v_2 M_2` ...(1) Also, if `v` is the total number of moles of the mixture in the vessels, then we know, `v = v_1 + v_2` ...(2) Solving (1) and (2) for `v_1` and `v_2`, we get, `v_1 ((v M_2 - m))/(M_2 - M_1), v_2 = (m - v M_1)/(M_2 - M_1)` Therefore, we get `m_1 = M_1 . ((vM_2 - m))/(M_2 - M_1)` and `m_2 = M_2 ((m - v M_1))/(M_2 - M_1)` or, `(m_1)/(m_2) = (M_1)/(M_2) ((v M_2 - m))/((m - v M_1))` One can also express the above result in terms of the effective molecular weight `M` of the mixture, defined as, `M = (m)/(v) = m (RT)/(p V)` Thus, `(m_1)/(m_2) = (M_1)/(M_2).(M_2 - M)/(M - M_1) = (1- M//M_2)/(M//M_1 - 1)` Using the date and table, we get : `M = 3.0 g` and, `(m_1)/(m_2) = 0.50`.

Discussion

3763.	A vessel of volume `V = 30 1` contains ideal gas at the temperature `0 ^@ C`. After a portion of the gas has been let out, the pressure in the vessel decreased by `Delta p = 0.78 atm` (the temperature remaining constant). Find the mass of the released gas. The gas density under the normal conditions `rho = 1.3 g//1`.
Answer» Let `m_1` and `m_2` be the masses of the gas in the vessel before and after the gas is released. Hence mass of the gas released, `Delta m = m_1 - m_2` Now from ideal gas equation `p_1 V = m_1 ( R)/(M) T_0` and `p_2 V = m_2 ( R)/(M) T_0` as `V` as `T` are same before and after the release of the gas. so, `(p_1 - p_2) V = (m_1 - m_2) ( R)/(M)T_0 = Delta m ( R)/(M) T_0` or, `Delta m=((p_1 - p_2)VM)/(R T_0) = (Delta p VM)/(R T_0)`....(1) We also know `p = rho (R)/(M) T so, (M)/(R T_0) = (rho)/(p_0)`...(2) (where `p_0` = standard atmospheric pressure and `T_0 = 273 K`) From Eqs. (1) and (2) we get `Delta m = rho V (Delta p)/(p_0) = 1.3 xx 30 xx (0.78)/(1) = 30 g`.

Discussion

3764.	A 2.5 kg cast iron pot contains 1.8 L of water at 20oC. How many minutes will it take to boil half away on a 5 kW burner with 40% heat loss. Specific heat of cast iron is 0.46 kJ/kg-K.
Answer» using formula p x t = ms\(\Delta\)t 5 x 10³ x t = 2.5 x 0.46 x [373 - 293] 5 x 10³ x t = 92 t = \(\frac{92}{5\,\times\,10^3}\) t = 18.4 x 10^-3 sec t = 0.3066 x 10^-3 m

3764.

A 2.5 kg cast iron pot contains 1.8 L of water at 20oC. How many minutes will it take to boil half away on a 5 kW burner with 40% heat loss. Specific heat of cast iron is 0.46 kJ/kg-K.

Answer»

using formula p x t = ms\(\Delta\)t

5 x 10³ x t = 2.5 x 0.46 x [373 - 293]

5 x 10³ x t = 92

t = \(\frac{92}{5\,\times\,10^3}\)

t = 18.4 x 10^-3 sec

t = 0.3066 x 10^-3 m

Discussion

3765.	An ideal gas is taken around ABCA as shown in the above P - V diagram. The work done during a cycle is A. ZeroB. `1/2PV`C. p PVD. PV
Answer» Correct Answer - D

Discussion

3766.	In some process the temperature of a substance depends on its entropy `S` as `T = a S^n`, where `a` and `n` are constants. Find the corresponding heat capacity `C` of the substance as a function of `S`. At what condition is `C lt 0` ?
Answer» Here `T = a S^n` or ` S = ((T)/(a))^((1)/(n))` Then `C = T(1)/(n) (T^((1)/(n) -1))/(a^(1//n)) = (S)/(n)` Clearly `C lt 0` if `n lt 0`.

Discussion

3767.	Find the entropy increment of an aluminium bar of mass `m = 3.0 kg` on its heating from the temperature `T_1 = 300 K` up to `T_2 = 600 K` if in this temperature interval the specific heat capacity of aluminimum varies as `c = a + bT`, where `a = 0.77 J//(g.K), b = 0.46 mJ//(g. k^2)`.
Answer» `Delta S = int_(T_1)^(T_2) (CdT)/(T) = int_(T_1)^(T_2) (m(a + bT))/(T) dT = mb (T_2 - T_1) + ma 1n (T_2)/(T_1)`.

Discussion

3768.	100 gram of ice at `0^@C` is converted into water vapour at `100^@C` Calculate the change in entropy.A. `-4.5 cal//K`B. `+4.5 cal//K`C. `+5.4 cal//K`D. `-5.4 cal//K`
Answer» Correct Answer - B

Discussion

3769.	Water of mass `m = 1.00 kg` is heated from the temperature `t_1 = 10^@C` up to `t_2 = 100 ^@C` at which it evaporated completely. Find the entropy increment of the system.
Answer» `Delta S = int _(T_1)^(T_2) mc (dT)/(T) + (mq)/(T_2) ~= m (c 1n (T_2)/(T_1)+(q)/(T_2))` =`10^3(4.18 1n (373)/(283) + (2250)/(373)) ~~ 7.2 kJ//K`.

Discussion

3770.	Enthalpy of neutralization of `H_(3)PO_(3) "with " NaOH " is" -106.68 "kJ"//"mol"`. If enthalpy of neutralization of HCL with NaOH is -55.84`"kJ"//"mole"`, then calculate enthalpy of ionization of `H_(3)PO_(3)` in to its ions in kJ.
Answer» Correct Answer - 5

Discussion

3771.	Enthalpy of neutralization of `H_(3)PO_(3) "with " NaOH " is" -106.68 "kJ"//"mol"`. If enthalpy of neutralization of HCL with NaOH is -55.84`"kJ"//"mole"`, then calculate enthalpy of ionization of `H_(3)PO_(3)` in to its ions in kJ.A. 50.84 kJ/molB. 5 kJ/molC. 2.5 kJ/molD. None of these
Answer» Correct Answer - B `H_(3)PC_(3)rarr2H^(+)+HPO_(3)^(2-) , Delta_(r )H=?` `2H^(+)+2OH^(-)rarr2H_(2)O`, `Delta_(r )H= -55.84xx2= -111.68` `-106.68=Delta _("ion")-55.84xx2` `Delta H_("ion")H=5 kJ//mol`.

Discussion

3772.	Which of the following is an extensive property?A. EthanlpyB. ConcentrationC. DensityD. Visocity
Answer» Enthalpy depends upon the quantity of substance.

Discussion

3773.	If temperature of the system remains constant during the course of change, the change isA. IsothermalB. AdiabaticC. IsobaricD. Isochroic
Answer» In isothermal process, temperature remains constant `(DeltaT = 0)`.

Discussion

3774.	The free energy change for a reversible reaction at equilibrium is:A. PositiveB. NegativeC. ZeroD. Can not say
Answer» Correct Answer - (C)

Discussion

3775.	`150 mL` of `0.5N HCl` solution at `25^(@)C` was mixed with `150 mL` of `0.5 N NaOH` solution at same temperature. Calculate the heat of neutralization of `HCl` with `NaOH`, if find temperature was recorded to be `29^(@)C`. `(rho_(H_(2)O)=1g//mL)`
Answer» Correct Answer - `-16 kcal`.

Discussion

3776.	The standard enthalpy of neutralization of `KOH` with `HCN` in dilute solution is `-2480 cal.mol^(-1)` and `-13.68 kcalmol^(-1)` respectively. Find the enthalpy of dissocitation of `HCN` at the same temperature.
Answer» Correct Answer - `11.2 Kcal`

Discussion

3777.	In which of the following process. the process is always non-feasible? (a) ∆H > O, ∆S > O (b) ∆H < O, ∆S > O(c) ∆H > O, ∆S < O(d) ∆H < O, ∆S > O
Answer» (c) ∆H > O, ∆S < O

Discussion

3778.	The heat of combustion of ethylene at `18^(@)C` and at constant volume is `-330.0 kcal` when water is obtained in liquid state. Calculate the heat of combustion at constant pressure and at `18^(@)C`?
Answer» The chemical equation for the combustion of `C_(2)H_(4)` is `{:(C_(2)H_(4)(g)+,3O_(2)(g)=,2CO_(2)(g)+,2H_(2)O(l),DeltaU^(Theta) =- 330.8kcal,),(1mol,3mol,2mol,,):}` Number of moles of reactants `= (1+3) = 4` Number of moles pf products `= 2` So, `Deltan = (2-4) =- 2` Given `DeltaU^(Theta) =- 330.0 kcal, Deltan =- 2, R = 2 xx 10^(-3) kcal` and `T = (18 +273) = 291 K` Applying `DeltaH^(Theta) = DeltaU^(Theta) + DeltanRT` `=- 330.0 + (-2) (2 xx 10^(-3)) (291)` `=- 331.164 kcal`

Discussion

3779.	For the reaction at `298 K` `2A+B rarr C` `DeltaH=400 kJ mol^(-1)` and `DeltaS=0.2 kJ K^(-1) mol^(-1)` At what temperature will the reaction becomes spontaneous considering `DeltaH` and `DeltaS` to be contant over the temperature range.
Answer» Correct Answer - 2000K From the expression, `DeltaG=DeltaH-TDeltaS` Assumingb the reaction at equilibrium , `DeltaT` for the reaction would be : `T=(DeltaH-DeltaG)(1)/(DeltaS)` `(DeltaH)/(DeltaS)(DeltaG=0` at equilibrium ) `=(400 kJ "mol"^(-1))/(0.2 kJ^(-1)"mol"^(-1))` T=2000 K For the reaction to be spontaneous, `DeltaG` must be negative . Hence , for the given reaction to be spontaneous, T should be greater than 2000 K.

Discussion

3780.	What is the normal boiling point of mercury? Given : `DeltaH _(f)^(@)(Hg,l)=0,S^(@)(Hg,l)=77.4 J//K-"mol"` `DeltaH _(f)^(@)(Hg,g)= 60.8 kJ//"mol", S^(@)(Hg, g)=174.4 J//K-"mol"`A. 624.8 KB. 626.8 KC. 636.8 KD. None of these
Answer» Correct Answer - B `HgiffHg(g)`, `Delta_(r )S^(@)=174.4-77.4 = 97 J//K-mol` `because" "DeltaG^(@)=DeltaH^(@)-T.DeltaS^(@)=0` `" "T=(DeltaH^(@))/(DeltaS^(@))` `" "=(60.8 xx1000)/(97)=626.8K`

Discussion

3781.	Change in Gibbs free energy is given by (a) ∆G = ∆H +T∆S (b) ∆G = ∆H – T∆S (c) ∆G = H x T∆S (d) none of these
Answer» (b) ∆G = ∆H – T∆S

Discussion

3782.	For an isolated system, `DeltaU=0`, what will be `Delta S`?
Answer» Correct Answer - `DeltaSgt0` `DeltaS` will be positive i.e., greater than zero Since `DeltaU=0,DeltaS` will be positive and the reaction will be spontaneous.

Discussion

3783.	The heat of reaction of `A+(1)/(2)O_(2)rarr AO` is -50 K call and `AO+(1)/(2)O_(2)rarr AO_(2)` is 100 Kcal. The heat of reaction for `A+O_(2)rarr AO_(2)` is :-A. `-50` K cal.B. `+50` K calC. 100 K cal.D. 150 K cal.
Answer» Correct Answer - B

Discussion

3784.	For isothermal expansion in case of an ideal gas `:`A. `DeltaG = DeltaS`B. `DeltaG = DeltaH`C. `DeltaG = - T. DeltaS`D. None of these
Answer» Correct Answer - C For ideal gas isothermal expansion `DeltaH = 0` `therefore" "DeltaG=-T.DeltaS` where `" "DeltaS=nRln((V_(2))/(V_(1)))`

Discussion

3785.	`C(s)+O_(2)(g)rarr CO_(2)(g)+94.0` K cal. `CO(g)+(1)/(2)O_(2)(g)rarr CO_(2)(g), 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)+(1)/(2)O_(2)(g)rarr CO(g)`A. `20.6`B. `26.3`C. `44.2`D. `161.6`
Answer» Correct Answer - B

Discussion

3786.	For isothermal expansion in case of an ideal gas `:`A. `DeltaG = DeltaS`B. `DeltaG = DeltaH`C. `DeltaG =-DeltaT. DeltaS`D. None of these
Answer» Correct Answer - C

Discussion

3787.	`C(s)+O_(2)(g)toCO_(2)(g)," "DeltaH=-94.3 kcal//mol` `CO(g)+(1)/(2)O_(2)(g)toCO_(2)(g)," "DeltaH=-67.4 kcal//mol` `O_(2)(g)to2O(g)," "DeltaH=117.4 kcal//mol` `CO(g)toC(g)+O(g)," "DeltaH=230.6 kcal//mol` Calculate `DeltaH` for `C(s)toC(g)` in `kcal//mol`A. 171B. 154C. 117D. 145
Answer» Correct Answer - D

Discussion

3788.	Which of the following reactions represents the enthalpy of formation of water?A. `H^(+)(aq) + OH^(-)(aq) rarr H_(2)O(l)`B. `H_(2)(g) + 1/2 O_(2)(g) rarr H_(2)O(l)`C. `2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l)`D. `2H^(+)(aq) + 2OH^)(-)(aq) rarr 2H_(2)O(l)`
Answer» Correct Answer - B

Discussion

3789.	Heat of neutralisation of a strong dibasic acid in dilute solution by NaOH is nearly :A. `-27.4` kcal/eqB. 13.7 kcal/molC. `-13.7`kcal/eqD. `-13.7` kcal/mol
Answer» Correct Answer - C

Discussion

3790.	Which of the following is not a closed system (i) Recket engine during dropulsion (ii) Pressure cooker (iii) Tea placed in a stell kettle Jet engineA. (i),(ii),(iii),(iv)B. (ii),(iii),(iv)C. (i),(ii),(iv)D. (i),(ii),(iii)
Answer» Correct Answer - C Tea placed in a steel kettle is a closed system because it can lose energy to the surroundings but not matter.

Discussion

3791.	The enthalpy of formation of `H_(2)O(l)` is -280.70 kJ/mol and enthalpy of neutralisation of a strong acid and strong base is -56.70 kJ/mol. What is the enthalpy of formation of `OH^(-)` ions?A. `-22.9` kJ/molB. `-224` kJ/molC. `58.7` kJ/molD. `214` kJ/mol
Answer» Correct Answer - B

Discussion

3792.	In a reversible process, `Delta S_(sys) + Delta S_(surr)` isA. `gt 0`B. `lt 0`C. `ge 0`D. `= 0`
Answer» Correct Answer - D "Reversible" under strict thermodynamics sense is a special way of carrying out a porcess such that the system is at times in perfect equilibrium with its surroundings. Thus, accoridng to the second law of thermodynamics, `Delta S_(uni) = Delta S_(sys) + Delta S_(surr) = 0`

Discussion

3793.	An ideal gasa undergoes through folowing cyclic procees. 1-2 : Reversible adialbatic comparession from `P_(1)V_(1)T_(1)"to"P_(2)V_(2)T_(2)` 2-3 : Reversible isochoric heating from `P_(2)V_(2)T_(2)" to " P_(3)V_(3)T_(3)` 3-4 Reversible adialbatic expansion from `P_(3)V_(3)T_(3)" to " P_(4)V_(4)T_(4)` 4-1 Reversible isochoric cooling from `P_(4)V_(4)T_(4)" to " P_(1)V_(1)T_(1)` Eefficiencyof the cycle is :A. `((T_(4)-T_(1))/(T_(3)-T_(2)))`B. `1-T_(1)/(T_(2))`C. `1-(V_(2)/(V_(1)))^(gamma-1)`D. `1-\|Q_(4-1)/(Q_(2-3))\|`
Answer» Correct Answer - a,b,c,d

Discussion

3794.	Which of the following is not the unit of energy?A. wattB. jouleC. calorieD. erg
Answer» Correct Answer - A Watt is a measure of the quantity of energy used per unit time (i.e., power) and equals 1 joule per second. The `SI` unit fo energy, kg `m^(2) s^(-2)`, is given the name joule `(J)` (pronounced jewl), after the English scientist James Prescott Juole who studies the energy concept. Joule is an extermely small unit. Calorie (cal) is a non- `SI` unit of energy, commonly used by chemist, originally defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius. This is only an approximate definition because the energy needed to heat water depends slightly on the temperature of water. Exact definition is `1 cal = 4.184 J`. Erg is a unit of energy (work in the dyne acting through a distance of 1 centimeters `(1 erg = 10^(-7) J)`. `1 cal gt 1 J gt 1 erg`

Discussion

3795.	For a spontaneous process, which of the following is always true?A. `DeltaG gt 0`B. `Delta_("total")S lt 0`C. `T DeltaS gt 0`D. `DeltaG lt 0`
Answer» `DeltaG lt 0` is always true for spontaneous process.

Discussion

3796.	The heats of combustion of `C_(2)H_(4), C_(2)H_(6)` and `H_(2)` gases are -1409.5 KJ, -1558.3 KJ and -285.6 KJ respectively. The heat of hydrogenation of ethene isA. `-136.8 KJ`B. `-13.68 KJ`C. `273.6 KJ`D. `1.368 KJ`
Answer» Correct Answer - A `C_(2)H_(4)+H_(2) rarr C_(2)H_(6)` `Delta H = H_(P) - H_(R)`

Discussion

3797.	The heats of combustion of `C_(2)H_(4), C_(2)H_(6)` and `H_(2)` gases are -1409.5 KJ, -1558.3 KJ and -285.6 KJ respectively. The heat of hydrogenation of ethene isA. `-136.8 KJ`B. `-13.68 KJ`C. 273.6 KJD. 1.368 KJ
Answer» Correct Answer - A `C_(2)H_(4)+H_(2)rarr C_(2)H_(6) , Delta H_("hydrogeneration")=?` `Delta H=Sigma (Delta H_("comb"))_(R )-Sigma (Delta H_("comb"))_(P)`

Discussion

3798.	Which one of the following has the same value as `Delta_(f)H^(Theta), CO`?A. `(1)/(2)Delta_(f)H^(Theta)(CO_(2))`B. `(1)/(2)Delta_(c)H^(Theta)("graphite")`C. `Delta_(f)H^(Theta) (CO_(2)) -Delta_(f)H^(Theta) ("graphite)"`D. `Delta_(c)H^(Theta) ("graphite") - Delta_(c)H^(Theta) (CO)`
Answer» `{:(C(g)+(1)/(2)O_(2)(g) rarr CO(g),,Delta_(f)HCO(g) =x),(i.C(g)+O_(2)(g)rarrCO_(2)(g),,DeltaH_(c)C(g)=y),(ii.CO(g)+(1)/(2)O_(2)(g)rarr CO_(2)(g),,DeltaH_(c)CO(g)=z):}` Substracting equation (ii) from (i), we get `{:(C(g)+(1)/(2)O_(2)(g)rarrCO(g),,DeltaH=y-z):}`

Discussion

3799.	For the reaction `N_(2) + 3 H_(2) = 2 NH_(3)`,A. `Delta U + 2 RT`B. `Delta U - 2RT`C. `Delta U + RT`D. `Delta U - RT`
Answer» Correct Answer - B According to thermodynamics, `Delta H = Delta U + Delta n_(g) RT` where `Delta n_(g) = sum n_(p) (g) - sum n_(R) (g)` `= - (2) - (1 + 3) - 2` `:. Delta H = Delta U - 2RT`

Discussion

3800.	Use bond energies to estimate the value of `DeltaH^(@)` for the reaction : `N_(2)(g)+3H_(2)(g)to2NH_(3)(g)` `{:(Bond,"EnergiesKJ/mol"),(H-H,436),(H-N,386),(N-N,193),(N=N,418),(N-=N,941):}`A. `-995KJ`B. `-590KJ`C. `-67Kj`D. `815KJ`
Answer» Correct Answer - c

Discussion

Explore topic-wise InterviewSolutions in .

Which one is the correct unit for entropy? (a) KJ mol (b) JK-1 mol (c) JK-1 mol-1 (d) KJ mol-1

30.4 KJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK-1 mol-1 Calculate the melting point of sodium chloride.

Calculate the equilibrium constant for the reaction given below at `400K`, if `DeltaH^(Theta) = 77.2 kJ mol^(-1)` and `DeltaS^(Theta) = 122 J K^(-1) mol^(-1)` `PCI_(5)(g) rarr PCI_(3)(g) +CI_(2)(g)`

Calculate the equilibrium constnat for the reaction given below at `400K`, if `DeltaH^(Theta) = 77.2 kJ "mole"^(-1)` and `DeltaS^(Theta) = 122 J K^(-1) "mole"^(-1)`. `PCI_(5)(g) rarr PCI_(3)(g) +CI_(2)(g)`

The temperature in K at which `DeltaG=0` for a given reaction with `DeltaH=-20.5 kJ mol^(-1)` and `DeltaS=-50.0 JK^(-1) mol^(-1)` isA. `-410`B. `410`C. `2.44`D. `-2.44`

Calculate the value of `DeltaG` at `700K` for the reaction, `nX to mB`. Given that value of `DeltaH=-113kJ mol^(-1)` and `DeltaS=-145JK^(-1)mol^(-1).`

If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride vapor, thenA. `Delta H = Delta U`B. `Delta H gt Delta U`C. `Delta H lt Delta U`D. there is no relationship

In thermodynamics, a process is called reversible whenA. the system changes into the surrounding spontaneouslyB. the surrounding are always in equilibrium with the systemC. there is no boundary between and surroundingD. surroundings and system change into each other

If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporization of 1 mol of water at 1 bar and `100^(@)`C is 41 kJ `mol^(-1)`. Calculate the internal energy, when 1 mol of water is vapourised at one bar pressure and `100^(@)`C.

Which among the following properties is/are extenstive properties?A. MoleB. Heat capacityC. Molar enthalpyD. Entropy

A liquid has a vapour pressure of 40mm Hg at `19.0^(@)C` and a normal boiling point of `78.3^(@)C` What is its enthalpy of vaporization in KJ .`mol^(-1)`?A. 42.4B. 18.4C. `5.10`D. 1.45

A vessel of volume `V = 20 1` contains a mixture of hydrogen and helium at a temperature `t = 20 ^@C` and pressure `p = 2.0 atm`. The mass of the mixture is equal is equal to `m = 5.0 g`. Find the ratio of the mass of hydrogen to that of helium in the given mixture.

A 2.5 kg cast iron pot contains 1.8 L of water at 20oC. How many minutes will it take to boil half away on a 5 kW burner with 40% heat loss. Specific heat of cast iron is 0.46 kJ/kg-K.

An ideal gas is taken around ABCA as shown in the above P - V diagram. The work done during a cycle is A. ZeroB. `1/2PV`C. p PVD. PV

In some process the temperature of a substance depends on its entropy `S` as `T = a S^n`, where `a` and `n` are constants. Find the corresponding heat capacity `C` of the substance as a function of `S`. At what condition is `C lt 0` ?

Find the entropy increment of an aluminium bar of mass `m = 3.0 kg` on its heating from the temperature `T_1 = 300 K` up to `T_2 = 600 K` if in this temperature interval the specific heat capacity of aluminimum varies as `c = a + bT`, where `a = 0.77 J//(g.K), b = 0.46 mJ//(g. k^2)`.

100 gram of ice at `0^@C` is converted into water vapour at `100^@C` Calculate the change in entropy.A. `-4.5 cal//K`B. `+4.5 cal//K`C. `+5.4 cal//K`D. `-5.4 cal//K`

Water of mass `m = 1.00 kg` is heated from the temperature `t_1 = 10^@C` up to `t_2 = 100 ^@C` at which it evaporated completely. Find the entropy increment of the system.

Enthalpy of neutralization of `H_(3)PO_(3) "with " NaOH " is" -106.68 "kJ"//"mol"`. If enthalpy of neutralization of HCL with NaOH is -55.84`"kJ"//"mole"`, then calculate enthalpy of ionization of `H_(3)PO_(3)` in to its ions in kJ.

Enthalpy of neutralization of `H_(3)PO_(3) "with " NaOH " is" -106.68 "kJ"//"mol"`. If enthalpy of neutralization of HCL with NaOH is -55.84`"kJ"//"mole"`, then calculate enthalpy of ionization of `H_(3)PO_(3)` in to its ions in kJ.A. 50.84 kJ/molB. 5 kJ/molC. 2.5 kJ/molD. None of these

Which of the following is an extensive property?A. EthanlpyB. ConcentrationC. DensityD. Visocity

If temperature of the system remains constant during the course of change, the change isA. IsothermalB. AdiabaticC. IsobaricD. Isochroic

The free energy change for a reversible reaction at equilibrium is:A. PositiveB. NegativeC. ZeroD. Can not say

`150 mL` of `0.5N HCl` solution at `25^(@)C` was mixed with `150 mL` of `0.5 N NaOH` solution at same temperature. Calculate the heat of neutralization of `HCl` with `NaOH`, if find temperature was recorded to be `29^(@)C`. `(rho_(H_(2)O)=1g//mL)`

The standard enthalpy of neutralization of `KOH` with `HCN` in dilute solution is `-2480 cal.mol^(-1)` and `-13.68 kcalmol^(-1)` respectively. Find the enthalpy of dissocitation of `HCN` at the same temperature.

In which of the following process. the process is always non-feasible? (a) ∆H &gt; O, ∆S &gt; O (b) ∆H &lt; O, ∆S &gt; O(c) ∆H &gt; O, ∆S &lt; O(d) ∆H &lt; O, ∆S &gt; O

The heat of combustion of ethylene at `18^(@)C` and at constant volume is `-330.0 kcal` when water is obtained in liquid state. Calculate the heat of combustion at constant pressure and at `18^(@)C`?

For the reaction at `298 K` `2A+B rarr C` `DeltaH=400 kJ mol^(-1)` and `DeltaS=0.2 kJ K^(-1) mol^(-1)` At what temperature will the reaction becomes spontaneous considering `DeltaH` and `DeltaS` to be contant over the temperature range.

What is the normal boiling point of mercury? Given : `DeltaH _(f)^(@)(Hg,l)=0,S^(@)(Hg,l)=77.4 J//K-"mol"` `DeltaH _(f)^(@)(Hg,g)= 60.8 kJ//"mol", S^(@)(Hg, g)=174.4 J//K-"mol"`A. 624.8 KB. 626.8 KC. 636.8 KD. None of these

Change in Gibbs free energy is given by (a) ∆G = ∆H +T∆S (b) ∆G = ∆H – T∆S (c) ∆G = H x T∆S (d) none of these

For an isolated system, `DeltaU=0`, what will be `Delta S`?

The heat of reaction of `A+(1)/(2)O_(2)rarr AO` is -50 K call and `AO+(1)/(2)O_(2)rarr AO_(2)` is 100 Kcal. The heat of reaction for `A+O_(2)rarr AO_(2)` is :-A. `-50` K cal.B. `+50` K calC. 100 K cal.D. 150 K cal.

For isothermal expansion in case of an ideal gas `:`A. `DeltaG = DeltaS`B. `DeltaG = DeltaH`C. `DeltaG = - T. DeltaS`D. None of these

`C(s)+O_(2)(g)rarr CO_(2)(g)+94.0` K cal. `CO(g)+(1)/(2)O_(2)(g)rarr CO_(2)(g), 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)+(1)/(2)O_(2)(g)rarr CO(g)`A. `20.6`B. `26.3`C. `44.2`D. `161.6`

For isothermal expansion in case of an ideal gas `:`A. `DeltaG = DeltaS`B. `DeltaG = DeltaH`C. `DeltaG =-DeltaT. DeltaS`D. None of these

`C(s)+O_(2)(g)toCO_(2)(g)," "DeltaH=-94.3 kcal//mol` `CO(g)+(1)/(2)O_(2)(g)toCO_(2)(g)," "DeltaH=-67.4 kcal//mol` `O_(2)(g)to2O(g)," "DeltaH=117.4 kcal//mol` `CO(g)toC(g)+O(g)," "DeltaH=230.6 kcal//mol` Calculate `DeltaH` for `C(s)toC(g)` in `kcal//mol`A. 171B. 154C. 117D. 145

Which of the following reactions represents the enthalpy of formation of water?A. `H^(+)(aq) + OH^(-)(aq) rarr H_(2)O(l)`B. `H_(2)(g) + 1/2 O_(2)(g) rarr H_(2)O(l)`C. `2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l)`D. `2H^(+)(aq) + 2OH^)(-)(aq) rarr 2H_(2)O(l)`

Heat of neutralisation of a strong dibasic acid in dilute solution by NaOH is nearly :A. `-27.4` kcal/eqB. 13.7 kcal/molC. `-13.7`kcal/eqD. `-13.7` kcal/mol

Which of the following is not a closed system (i) Recket engine during dropulsion (ii) Pressure cooker (iii) Tea placed in a stell kettle Jet engineA. (i),(ii),(iii),(iv)B. (ii),(iii),(iv)C. (i),(ii),(iv)D. (i),(ii),(iii)

The enthalpy of formation of `H_(2)O(l)` is -280.70 kJ/mol and enthalpy of neutralisation of a strong acid and strong base is -56.70 kJ/mol. What is the enthalpy of formation of `OH^(-)` ions?A. `-22.9` kJ/molB. `-224` kJ/molC. `58.7` kJ/molD. `214` kJ/mol

In a reversible process, `Delta S_(sys) + Delta S_(surr)` isA. `gt 0`B. `lt 0`C. `ge 0`D. `= 0`

Which of the following is not the unit of energy?A. wattB. jouleC. calorieD. erg

For a spontaneous process, which of the following is always true?A. `DeltaG gt 0`B. `Delta_("total")S lt 0`C. `T DeltaS gt 0`D. `DeltaG lt 0`

The heats of combustion of `C_(2)H_(4), C_(2)H_(6)` and `H_(2)` gases are -1409.5 KJ, -1558.3 KJ and -285.6 KJ respectively. The heat of hydrogenation of ethene isA. `-136.8 KJ`B. `-13.68 KJ`C. `273.6 KJ`D. `1.368 KJ`

The heats of combustion of `C_(2)H_(4), C_(2)H_(6)` and `H_(2)` gases are -1409.5 KJ, -1558.3 KJ and -285.6 KJ respectively. The heat of hydrogenation of ethene isA. `-136.8 KJ`B. `-13.68 KJ`C. 273.6 KJD. 1.368 KJ

Which one of the following has the same value as `Delta_(f)H^(Theta), CO`?A. `(1)/(2)Delta_(f)H^(Theta)(CO_(2))`B. `(1)/(2)Delta_(c)H^(Theta)("graphite")`C. `Delta_(f)H^(Theta) (CO_(2)) -Delta_(f)H^(Theta) ("graphite)"`D. `Delta_(c)H^(Theta) ("graphite") - Delta_(c)H^(Theta) (CO)`

For the reaction `N_(2) + 3 H_(2) = 2 NH_(3)`,A. `Delta U + 2 RT`B. `Delta U - 2RT`C. `Delta U + RT`D. `Delta U - RT`

Use bond energies to estimate the value of `DeltaH^(@)` for the reaction : `N_(2)(g)+3H_(2)(g)to2NH_(3)(g)` `{:(Bond,"EnergiesKJ/mol"),(H-H,436),(H-N,386),(N-N,193),(N=N,418),(N-=N,941):}`A. `-995KJ`B. `-590KJ`C. `-67Kj`D. `815KJ`

In which of the following process. the process is always non-feasible? (a) ∆H > O, ∆S > O (b) ∆H < O, ∆S > O(c) ∆H > O, ∆S < O(d) ∆H < O, ∆S > O