4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 30 InterviewSolution

1451.	Give examples of endothermic and exothermic reactions.
Answer» Reaction in which heat is given out along with the products is called exothermic reaction: Example: 1. Burning of methane gas releases a large amount of energy. Hence it is an exothermic reaction. 2. Reaction of calcium oxide and water Reactions that absorb energy or require energy in order to proceed are called endothermic reactions. Example : (a) When ammonium chloride is dissolved in water in a test tube, the test tube becomes cold. (b) Reaction of barium hydroxide with ammonium chloride.

Discussion

1452.	Heat of neutralisation of strong acid and strong base is constant. Why?
Answer» The enthalpy of neutralization of all strong acids and strong bases is always constant, i.e., -57.1 kJ. The explanation to this generalization can be provided on the basis of Arrhenius theory of ionization. The strong acids and strong bases are almost completely ionized in dilute aqueous solutions. The neutralization of strong acid and strong bases simply involve the combination of H⁺ ions (from acid) and OH^– ions (from base) to form water molecules. It has been found experimentally that when 1 mole of water is formed by the neutralization of 1 mole of H⁺ ions and 1 mol of OH^– in aqueous solutions, 57.1 kJ of energy is released.

Discussion

1453.	Which of the following reactions have maximum heat of neutralisation?(a) NH4 OH and CH3 COOH(b) NH4 OH and HCl(c) NaOH and CH3 COOH(d) NaOH and HCl
Answer» (d) NaOH and HCl

Discussion

1454.	A Carnot engine converts one-fifth of heat given into work. If temperature of sink is reduced by `80^(@)`, efficiency gets doubled. If temperature of sourceand sink is `T_(1)`and `T_(2)` respectively then calculate value of `(T_(1)-T_(2))/(10)`
Answer» Correct Answer - 8

Discussion

1455.	Define internal energy change.
Answer» Internal energy change:- If E₁ is the internal energy of the system in the initial state (that is, before the change) and E₂ is the internal energy of the system in the final state (after change), then the internal energy change (ΔE) is given by ΔE = E₂ - E₁ = E_final - E_initial

1455.

Define internal energy change.

Answer»

Internal energy change:- If E₁ is the internal energy of the system in the initial state (that is, before the change) and E₂ is the internal energy of the system in the final state (after change), then the internal energy change (ΔE) is given by

ΔE = E₂ - E₁ = E_final - E_initial

Discussion

1456.	With the help of thermochemical equation, calculate ΔfH° at 298 K for the following reactions: C(graphite) + O2(g) → CO2(g) ; ΔfH° = -393.5kJ/mol H2(g) + 1/2O2(g) → H2O(l) ; ΔfH° = -285.8kJ/mol CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ; ΔfH° = +890.3kJ/mol
Answer» C_(graphite) + O₂(g) → CO₂(g) ; Δ_fH° = -393.5 kJ/mol …(1) H₂(g) + 1/2O₂(g) → H₂O(l) ; Δ_fH° = -285.8 kJ/mol …(2) CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) Δ_fH° = +890.3 kJ/mol …(3) Here we want one mole of C(graphite) as reactant, so we write down equation (1) as such, we want two moles of H2(g) as reactant, so we multiply equation (2) by 2, we want C_(graphite) + O₂ (g) → CO₂(g) ; Δ_fH° = -393.5 kJ / mol …(1) 2H₂(g) + O₂(g) → 2H₂O(l) ; Δ_fH° = 2(-285.8 kJ/mol) …(2) CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ; Δ_fH° = +890.3 kJ/mol . … (3) Adding we obtain: C_(graphite)+ 2H₂(g) → CH₄(g); Δ_fH° = -74.8 kJ/mol

1456.

With the help of thermochemical equation, calculate ΔfH° at 298 K for the following reactions: C(graphite) + O2(g) → CO2(g) ; ΔfH° = -393.5kJ/mol H2(g) + 1/2O2(g) → H2O(l) ; ΔfH° = -285.8kJ/mol CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ; ΔfH° = +890.3kJ/mol

Answer»

C_(graphite) + O₂(g) → CO₂(g) ; Δ_fH° = -393.5 kJ/mol …(1)

H₂(g) + 1/2O₂(g) → H₂O(l) ; Δ_fH° = -285.8 kJ/mol …(2)

CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) Δ_fH° = +890.3 kJ/mol …(3)

Here we want one mole of C(graphite) as reactant, so we write down equation (1) as such, we want two moles of H2(g) as reactant, so we multiply equation (2) by 2, we want

C_(graphite) + O₂ (g) → CO₂(g) ; Δ_fH° = -393.5 kJ / mol …(1)

2H₂(g) + O₂(g) → 2H₂O(l) ; Δ_fH° = 2(-285.8 kJ/mol) …(2)

CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ; Δ_fH° = +890.3 kJ/mol . … (3)

Adding we obtain:

C_(graphite)+ 2H₂(g) → CH₄(g); Δ_fH° = -74.8 kJ/mol

Discussion

1457.	For a reaction :`2A(g) to B(g), DeltaH =-40 Kcal`. If rate constant for disapperance of A is `10^(-2) M^(-1) sec^(-1)` then identify the opations which will be correct [Assume gases to behave ideally and reaction to be occuring at 300 K] [Take : R=2 cal/mole K]A. The reaction must be elementary reaction.B. Change in internal energy of the reaction will be `-19.7 kcal//"mole"` ofA consumed.C. The rate at which heat will be liberated initially when 2 moles of A is taken in 3 litre rigid vessel will be approx `0.26 kcal//sec`.D. Time requried for concentration of A
Answer» Correct Answer - b,d

Discussion

1458.	If the heats of formation of `C_(2)H_(2)` and `C_(6)H_(6)` are `230 kJ mol^(-1)` and `85 kJ mol^(-1)` respectively, the `DeltaH` value for the trimerisation of `C_(2)H_(2)` isA. `-605 kJ`B. `-205 kJ`C. `205 kJ`D. `605 kJ`
Answer» Correct Answer - A `3C_(2)H_(2) rarr C_(6)H_(6) , DeltaH=H_(P)-H_(R)`

Discussion

1459.	The expansion of a gas into an evacuated space takes place non-spontaneously. A process in which all steps cannot be retraced by themselves is called a spontaneous process.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-18B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-18C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE
Answer» Correct Answer - D

Discussion

1460.	A gobar gas plant produces methane and supplies to the families. If a family requires 15,000 kJ of energy per day, how many days will the quote of 10 kg will last ? `(Delta_(c)H "of" CH_(4) = - "1665 kJ "mol^(-1))`
Answer» Correct Answer - 69.37 days

Discussion

1461.	An athlete is given 180 g of glucose `(C_(6)H_(12)O_(6))`. He utilises 50% of the energy due to internal combustion in the body. In order to avoid storage of energy in the body, calculate the masss of water he would need to perspire. Given enthalpy of combustion of glucose is -2800 kJ `mol^(-1)` and enthalpy of evaporation of water is 44 kJ `mol^(-1)`
Answer» Correct Answer - 572.73 g

Discussion

1462.	90 g of water spilled out from a vessel in the room on the floor. Assuming that water vapour behaving as an ideal gas, calculate the internal energy change when the spilled water undergoes complete evaporation at `100^(@)C`. (Given the molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ `mol^(-1)`).
Answer» Correct Answer - 25.49 kJ `mol^(-1)`

Discussion

1463.	What is Extensive property? Give example.
Answer» It is a property of which depends on the amount of the substance present in the system. Example: Mass, Volume, Energy

1463.

What is Extensive property? Give example.

Answer»

It is a property of which depends on the amount of the substance present in the system.

Example: Mass, Volume, Energy

Discussion

1464.

Write any two differences between Isothermal and Adiabatic process.

Answer»

Isothermal Process	Adiabatic Process
1. Internal energy does not change.	1. Internal energy changes.
2. Heat change occurs between system and surrounding.	2. Heat does not change.

Discussion

1465.	A sample of an ideal monoatomic gas is taken round the cycle ABCA as shown in the figure the work done during the cycle is A. ZeroB. 2 PVC. 6 PVD. 9 PV
Answer» Correct Answer - B

Discussion

1466.	Assume ideal gas behavior for all the gases considered and vibratonal degree of freedom to be active. Separated equimolar sample of `Ne,O_(2), So_(2) and CH_(4)` were subjedted to a two step process as mentioned . Initiallyt all are at same state of temperture and pressure. Step -I: All undergo reversible abiabatic expansion to attain same final volume thereby causing the decreasae in their temperature. Step -II : After step I, all are given appropriate amount of heat isochorically to restore the original temperature . Mark the correct optoin(s).A. Due to step I only , the decrease in temperature will be maximum for Ne.B. During step II , heat given will be minimum for `CH_(4)`C. There will be no net change in internal energy for any of the gas after both the steps of process are completed.D. The `(@) P-V` graph of `CH_(4)` and `SO_(2)` wll be same for overall process.
Answer» Correct Answer - a,c

Discussion

1467.	A bomb calorimeter has a heat capacity of `783J xx^(@)c^(-1)` and contains 254 g of water which has a specific heat of `4.184Jxxg^(-1)xx^(@)C_(-1)` .How much heat is enolved or absorbed by a reaction when the temperature goes from `23.73^(@) "to" 26.01^(@)` C ?A. 1.78 KJ absorbedB. 2.42KJ absorbedC. 1.78 KJ evolvedD. 4.21KJ evolved
Answer» Correct Answer - d

Discussion

1468.	For which reaction from the following, `DeltaS` will be maximum?A. `Ca(s)+1//2O_(2)(g)rarr CaO(s)`B. `CaCO_(3)(s)rarr CaO(s)+CO_(2)(g)`C. `C(s)+O_(2)(g)rarr CO_(2)(g)`D. `N_(2)(g)+O_(2)(g)rarr 2NO(g)`
Answer» Correct Answer - B

Discussion

1469.	An adiabatic process is one in which :-A. Temperature of the system does not changeB. The system is not closed to heat transferC. There is no entropy changeD. None of these
Answer» Correct Answer - C

Discussion

1470.	Entropy meansA. DisordernessB. RanomnessC. OrdernessD. both 1 & 2
Answer» Correct Answer - D

Discussion

1471.	`Delta S` for the reaction , `MgCO_(3)(s)rarr MgO(s)+CO_(2)(g)` will be :
Answer» Correct Answer - C

Discussion

1472.	Change in entropy is negative forA. Bromine `(l) rarr` Bromine (g)B. `C(s)+H_(2)O(g)rarrCO(g)+H_(2)(g)`C. `N_(2)(g,10 atm)rarr N_(2)(g, 1 atm)`D. `Fe(at 400 K)rarr Fe(at 300 K)`
Answer» Correct Answer - D

Discussion

1473.	When two gases are mixed the entropy :-A. Remains constantB. DecreasesC. IncreasesD. Becomes zero
Answer» Correct Answer - C

Discussion

1474.	In which reaction `Delta S` is positive :-A. `H_(2)O(l)rarr H_(2)O(s)`B. `3O_(2)(g)rarr 2O_(3)(g)`C. `H_(2)O(l)rarr H_(2)O(g)`D. `N_(2)(g)+3H_(2)(g)rarr 2NH_(3)(g)`
Answer» Correct Answer - C

Discussion

1475.	When the egg is hard boiled, there isA. Increase in disorderB. Decrease in disorderC. No change in disorderD. `Delta G` is negative
Answer» Correct Answer - A

Discussion

1476.	The heat of combustion of `H_(2)(g)` at constant pressure and `300k` is `-280kJ mol^(-1)`. What will be the heat of combustion at constant volume and at `300K`?
Answer» `H_(2)(g) = (1)/(2)O_(2)(g) rarr H_(2)O(l) DeltaH =- 280 kJ mol^(-1)` `DeltaH = DeltaU +DeltanRT, Deltan = 0 - (3)/(2) =- (3)/(2)`, `DeltaH = DeltaU -(3)/(2) xx 8.314 xx 300` `- 280 kJ = DeltaU - (74826)/(20) xx (1)/(1000) kJ` `rArr DeltaU =- 276.2587 kJ`

Discussion

1477.	The standard absolute entropy of a substance, `(S^(Theta))` is the entropy of the substance in its standard state at `1` atm, temperature beingA. `0K`B. `298K`C. `398K`D. `273 K`
Answer» Correct Answer - `298 K`

Discussion

1478.	Heat of reaction for `C_(6)H_(12)O_(6(s))+6O_(2(g)) rarr 6CO_(2(g))+6H_(2)O_(v)` at constant pressure is `-651 kcal` at `17^(@)C`. Calculate the heat of reaction at constant volume at `17^(@)C`.
Answer» `DeltaH = DeltaU +DeltanRT….(i)` Given `DeltaH =- 651 xx10^(3) cal, R = 2 cal` Substituting all value in equation (i), we get `T = 290 K`, `Deltan = 6+6 -6 = 6` `:. =- 651 xx 10^(3) = DeltaU +6 xx 2xx 290` and `DeltaU =- 654480 cla or -654.48 kcal ` Note: For `C_(6)H_(12)O_(6)(g) +6O_(2)(g) rarr 6CO_(2)(g) +6H_(2)O(v) Deltan =` Number of products molecules -Number of reactants molecules (only in gaseous phase) `=6 +6 -6 = 6` Here, `H_(2)O(v)` is also in gaseous state.

Discussion

1479.	The enthalpy change of a reaction does not depand onA. State of reactants and productsB. Nature of reactants and productsC. Different intermediate reactionsD. Initial and final enthalpy change of reaction
Answer» Correct Answer - C

Discussion

1480.	What is Equation of state and examples ?
Answer» The mathematical relation between the pressure, volume and temperature of a thermodynamic system is called its equation of state. For example, the equation of state of n moles of an ideal gas can be written as PV = n RT

1480.

What is Equation of state and examples ?

Answer»

The mathematical relation between the pressure, volume and temperature of a thermodynamic system is called its equation of state.

For example, the equation of state of n moles of an ideal gas can be written as

PV = n RT

Discussion

1481.	What is meant by Thermodynamic equilibrium ?
Answer» A system is said to be in the state of thermodynamic equilibrium, if the microscopic variables of the describing the thermodynamic state of the system do not change with time. A system in a state of thermodynamic equilibrium possess mechanical, thermal and chemical equilibria simultaneously.

Discussion

1482.	Difference between heat and work.
Answer» (i) Heat is a mode of energy transfer due to temperature difference between the system and the surroundings. Work is the mode of energy transfer brought about by means that do not involve temperature difference such as moving the piston of a cylinder containing the gas, by raising or lowering the weight connected to it. (ii) When heat is supplied to a gas, its molecules move faster in all directions at random. So heat is a mode of energy transfer that produces random motion, when a piston compresses a gas to do work on it, it forces the molecules to move in the direction of piston’s motion. So work may be regarded as the mode of energy transfer that produces organised motion.

1482.

Difference between heat and work.

Answer»

(i) Heat is a mode of energy transfer due to temperature difference between the system and the surroundings. Work is the mode of energy transfer brought about by means that do not involve temperature difference such as moving the piston of a cylinder containing the gas, by raising or lowering the weight connected to it.

(ii) When heat is supplied to a gas, its molecules move faster in all directions at random. So heat is a mode of energy transfer that produces random motion, when a piston compresses a gas to do work on it, it forces the molecules to move in the direction of piston’s motion. So work may be regarded as the mode of energy transfer that produces organised motion.

Discussion

1483.	What is Zeroth Law of Thermodynamics ?
Answer» If systems A and B are each in thermal equilibrium with a third system separately C, then A and B are in thermal equilibrium with each other. The zeroth law leads to the concept of temperature.

Discussion

1484.	Define Thermodynamic variables. Give the state of Thermodynamic variables.
Answer» Pressure, volume, temperature, internal energy and the number of moles are called thermodynamic variables. For µ moles of an ideal gas, equation of state is PV = µRT. Thermodynamic State Variables : (i) Intensive variables (ii) Extensive variables.

1484.

Define Thermodynamic variables. Give the state of Thermodynamic variables.

Answer»

Pressure, volume, temperature, internal energy and the number of moles are called thermodynamic variables.

For µ moles of an ideal gas, equation of state is PV = µRT.

Thermodynamic State Variables :

(i) Intensive variables

(ii) Extensive variables.

Discussion

1485.	Enthalpy of neutralization is defined as the enthalpy change when `1` mole of acid/base is completely neutralized by base/acid in dilute solution. For strong acid and strong base neutralization net chemical change is `" "H^(+)(aq) + OH^(-)(aq) rarr H_(2)O(l), " " Delta_(r)H^(@) = -55.84 KJ//mol` `DeltaH_("ionization")^(@)` of aqueous solution of strong acid and strong base is zero. When a dilute solution of a weak acid or base is neutalized, teh enthalpy of neutralization is some what less because of the absorption of heat in the ionization of the weak acid or base, for weak acid/base `DeltaH_("neutrilization")^(2) = DeltaH_("ionization")^(@) + Delta_(r)H^(@) " " (H^(+) + OH^(-) rarr H_(2)O)` Under the same condition how many mL of `0.1` M NaOH and `0.05` M `H_(2)`A (strong diprotic acid) solution should be mixed for a total volume of `100` mL produce the highest rise in temperature:A. `25:75`B. `50:50`C. `75:25`D. `66.66:33.33`
Answer» Correct Answer - B For max. rise in temp. max. neutralization of `H^(+)` and `OH^(-)` required . If we take equal volume , all `H^(+) (5 m-"mole")` will react with all `OH^(-)(5 m-"mole").`

Discussion

1486.	Heat of dissociation of acetic acidis 0.30 kcal`mol^(-1)` . Hence,enthalpy change when 1 mol of`Ca(OH)_(2)` is completely neutralized by acetic acid would beA. `-13.4 kcal`B. `-27.1 kcal`C. `-26.8 kcal`D. `-27.4 kcal`
Answer» Correct Answer - c `Ca(OH)_(2) rarr Ca^(2+) + 2OH^(-) , i.e., 1` mole of`Ca(OH)_(2)` gives two moles of `OH^(-)` ions. Hence, when `Ca(OH)_(2)` is completely neutralized,heat produced `= 2 xx 13.7 kcal = 27.4 kcal` 2 moles of `OH^(-)` ions require 2 moles of `H^(+)` ions for complete neutralization. Heat of dissociation of acetic acidto produce 2 moles of`H^(+)` ions `= 2 xx 0.30 kcal = 0.6 kcal`. Hence, net heat produced `=27.4 - 0.6 =26.8 kcal, i.e., DeltaH = - 26.8 kcal`

Discussion

1487.	The enthalpy change when 1 L of 1 M `H _(2)SO_(4)` is completely neutralized by 1L of1 M `Ca(OH)_(2)` will beA. `-13.7 kcal`B. `-27.4 kcal`C. `-1.37 kcal`D. `- 2.74 kcal`
Answer» Correct Answer - b 1 L of 1 M `H_(2)SO_(4)` gives 2 moles of`H^(+)` ions 1L of`M Ca9OH)_(2)` gives 2 moles of `OH^(-) ` ions Hence, for complete neutralization of 2 molesof `H^(+)` ions by 2 molesof `OH^(-)` ions. `DeltaH = 2 ( - 13.7) kcal =- 27.4 kcal`

Discussion

1488.	The enthalpies of elements in their standard states are takeen as zero. The enthalpy of formation of a compoundA. is always negativeB. is always positiveC. may be positive or negativeD. is never negative
Answer» Correct Answer - c Combination of elements to form a compound can be exothermic or endothermic`( e.g., C + O_(2) rarrCO_(2)` is exothermic whereas `C + 2S rarrCS_(2)` is endothermic ). Hence, enthalpy of formation can be `+ve` or `-ve`.

Discussion

1489.	ΔU of combustion of methane is -x kJ mol-1. The value of ΔH° is(a) = ΔU°(b) > ΔU°(c) < ΔU°(d) = 0
Answer» Answer is (b) > ΔU°

Discussion

1490.	The enthalpies of all elements in their standard states are:(a) unity(b) zero(c) < 0(d) different for each element
Answer» Answer is (b) zero

Discussion

1491.	For the process to occur under adiabatic conditions, the correct condition is(a) ΔT = 0(b) Δp = 0(c) q = 0(d) W = 0
Answer» Answer is (c) q = 0

Discussion

1492.	A thermodynamic state function is a quantity:(a) used to determine heat changes(b) whose value is independent of path(c) used to determine pressure volume work(d) whose value depends on temperature only
Answer» Answer is (b) whose value is independent of path

Discussion

1493.	If we mix `H_(2)` and `O_(2)`, they do not combine to form `H_(2)O`, is the reaction spontaneous or non-spontaneous ? Give reason.
Answer» Reaction is spontaneous because it takes place when intiated by an electric spark.

Discussion

1494.	Calculate `Delta H ^(@)` for the reaction `CH_(2)= CH_(2) + 3O_(2) rarr 2CO_(2)+ 2H_(2)O` Given that the average bond energies of the different bonds are `:` `{:("Bond",C-H,O=O,C=O,O-H,C=C),("Bond energy " ( kJ mol^(-1)),414,499,724,460,619):}`
Answer» Correct Answer - `-964 Kj mol^(-1)` `Deta_(r)H [B.E. (C=C) + 4B.E. (C-H) + 3 B.E.(O-O) ] - [ 2 xx 2 xx B.E. ( C=O) + 2 xx 2 xx B.E.(O-H) ]` ` [619+ 4 ( 414)]+ 3 ( 499) ] - [ 4 ( 724)+ 4 ( 460 ) ] kJ = - 964 kJ`

Discussion

1495.	If an additional pressure `Delta p` of a saturated vapour over a convex spherical surface of a liquid is considerably less than the vapour pressure over a plane surface, then `Delta p = (rho_v//rho_l) 2 alpha//r`, where `rho_v` and `rho_l)` are the densities of the vapour and the liquid, `alpha` is the surface tension, and `r` is the radius of curvature of the surface. Using this formula, find the diameter of water droplets at which the saturated vapour pressure exceeds the vapour pressure over the plane surface by `eta = 1.0 %` at a temperature `t = 27 ^@C`. The vapour is assumed to be an ideal gas.
Answer» Given `Delta P = (rho_v)/(rho_l) (2 alpha)/(r) = (rho_v)/(rho_l) xx (4 alpha)/(d) = eta p_(vap) = eta ((m)/(M) RT)/(V_(vap)) = (eta RT)/(M) rho_v` or `d = (4 alpha M)/(rho_1 RT eta)` For water `alpha = 73 "dynes"//cm, M = 18 gm, rho_l = gm//c c, T = 300 K`, and with `eta ~~ 0.01`, we get `d ~~ 0.2 mu m`.

Discussion

1496.	Hydrazine , `N_(2)H_(4) ,` contains aN-N single bond and 4N -H bonds . Use bond energies to calculate `DeltaH`in KJ for the reaction : `N_(2)+2H_(2)toN_(2)H_(4)` `{:(Bond ,"Energies" (KJ.mol^(-1))),(H-H,436),(H-N,386),(N-N,193),(N=N,418),(N-=N,941):}`A. `-425KJ`B. `-76KJ`C. `76KJ`D. `245KJ`
Answer» Correct Answer - c

Discussion

1497.	Bomb calorimeters are the devices that are used to experimentally determine `DeltaH` and `DeltaU` of any reaction by measuring the temperature change in the water bath Hence , the experimatal data can be used for cross-checking theortical data, From the following theoretical data answer the question that follow: Thermodynamic data: `DeltaH_(f)^(@)Ch_(4)(g)=- 15 kcal //"mole", DeltaH_(f)^(@) CO_(2)(g)=-90 kcal//"mole",` `DeltaH_(f)^(@)H_(2)O(l)=-60 Kcal//"mole"` `S_(m)^(@) CH_(4)(g)= 40 cal//"mole Kelivn"` `S_(m)^(@) O(g)= 45 cal//"mole Kelivn"` `S_(m)^(@) H_(2)O(g)= 15 cal//"mole Kelivn"` `S_(m)^(@) CO_(2)(g)= 50 cal//"mole Kelivn"` Calorimeter data: (1) Water equivlent of calorimaeter = 36 gm (2) Sepcific heat capacity of water = `1 cal//gm .^(@)C` (2) mass of water inthe water bath =164 gm All data at 300 K, `R=2 cal//"mol "K` . If calorimeter is modified so as to convert chemical energy into electrical by ensuring that the reaction is occurring at constant pressure then how much electron work can be obtained by combustion of 0.1 mole of `CH_(4) (g)`?A. `19.5 kcal`B. `18 kcal`C. `19.62 kcal`D. `18.12 kcal`
Answer» Correct Answer - b

Discussion

1498.	Find the mass of all molecules leaving one square centimetre of water surface per second into a saturated water vapour above it at a temperature `t = 100 ^@C`. It is assumed that `eta = 3.6 %` of all water vapour molecules falling on the water surface are retained in the liquid phase.
Answer» In equilibrium the number of "liquid" molecules evaporoting must equals the number of "vapour" molecules condensing. By kinetic theory, this number is `eta xx (1)/(4) n lt v gt = eta xx (1)/(4) n xx sqrt((8)/(pi)(kT)/(m))` Its mass is `mu = m xx eta xx n xx sqrt((kT)/(2 pi m)) = eta n k T sqrt((m)/(2 pi kt))` `=eta p_0 sqrt((M)/(2 pi RT)) = 0.35 g//cm^2.s.` where `p_0` is atmospheric pressure and `T = 373 K` and `M` = molecular weight of water.

Discussion

1499.	Bomb calorimeters are the devices that are used to experimentally determine `DeltaH` and `DeltaU` of any reaction by measuring the temperature change in the water bath Hence , the experimatal data can be used for cross-checking theortical data, From the following theoretical data answer the question that follow: Thermodynamic data: `DeltaH_(f)^(@)Ch_(4)(g)=- 15 kcal //"mole", DeltaH_(f)^(@) CO_(2)(g)=-90 kcal//"mole",` `DeltaH_(f)^(@)H_(2)O(l)=-60 Kcal//"mole"` `S_(m)^(@) CH_(4)(g)= 40 cal//"mole Kelivn"` `S_(m)^(@) O(g)= 45 cal//"mole Kelivn"` `S_(m)^(@) H_(2)O(g)= 15 cal//"mole Kelivn"` `S_(m)^(@) CO_(2)(g)= 50 cal//"mole Kelivn"` Calorimeter data: (1) Water equivlent of calorimaeter = 36 gm (2) Sepcific heat capacity of water = `1 cal//gm .^(@)C` (2) mass of water inthe water bath =164 gm All data at 300 K, `R=2 cal//"mol "K` . 1. Calculate `DeltaU_("combustion")` of `CH_(4)(g)` at 300 K.A. `-195 kcal//"mole"`B. `-196.2 kcal//"mole"`C. `-193.8 kcal//"mole"`D. `-235 kcal//"mole"`
Answer» Correct Answer - c

Discussion

1500.	Bomb calorimeters are the devices that are used to experimentally determine `DeltaH` and `DeltaU` of any reaction by measuring the temperature change in the water bath Hence , the experimatal data can be used for cross-checking theortical data, From the following theoretical data answer the question that follow: Thermodynamic data: `DeltaH_(f)^(@)Ch_(4)(g)=- 15 kcal //"mole", DeltaH_(f)^(@) CO_(2)(g)=-90 kcal//"mole",` `DeltaH_(f)^(@)H_(2)O(l)=-60 Kcal//"mole"` `S_(m)^(@) CH_(4)(g)= 40 cal//"mole Kelivn"` `S_(m)^(@) O(g)= 45 cal//"mole Kelivn"` `S_(m)^(@) H_(2)O(g)= 15 cal//"mole Kelivn"` `S_(m)^(@) CO_(2)(g)= 50 cal//"mole Kelivn"` Calorimeter data: (1) Water equivlent of calorimaeter = 36 gm (2) Sepcific heat capacity of water = `1 cal//gm .^(@)C` (2) mass of water inthe water bath =164 gm All data at 300 K, `R=2 cal//"mol "K` . Calculate rise in temperature of calorimeter if 0.01 moles of `CH_(4)(g)` undergoes combustion in the above bonb calorimeter at 300 K .A. `9.18 K`B. `9.696 K`C. `12 K`D. `8.8 K`
Answer» Correct Answer - b

Discussion

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Give examples of endothermic and exothermic reactions.

Heat of neutralisation of strong acid and strong base is constant. Why?

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A Carnot engine converts one-fifth of heat given into work. If temperature of sink is reduced by `80^(@)`, efficiency gets doubled. If temperature of sourceand sink is `T_(1)`and `T_(2)` respectively then calculate value of `(T_(1)-T_(2))/(10)`

Define internal energy change.

With the help of thermochemical equation, calculate ΔfH° at 298 K for the following reactions: C(graphite) + O2(g) → CO2(g) ; ΔfH° = -393.5kJ/mol H2(g) + 1/2O2(g) → H2O(l) ; ΔfH° = -285.8kJ/mol CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ; ΔfH° = +890.3kJ/mol

If the heats of formation of `C_(2)H_(2)` and `C_(6)H_(6)` are `230 kJ mol^(-1)` and `85 kJ mol^(-1)` respectively, the `DeltaH` value for the trimerisation of `C_(2)H_(2)` isA. `-605 kJ`B. `-205 kJ`C. `205 kJ`D. `605 kJ`

A gobar gas plant produces methane and supplies to the families. If a family requires 15,000 kJ of energy per day, how many days will the quote of 10 kg will last ? `(Delta_(c)H "of" CH_(4) = - "1665 kJ "mol^(-1))`

What is Extensive property? Give example.

Write any two differences between Isothermal and Adiabatic process.

A sample of an ideal monoatomic gas is taken round the cycle ABCA as shown in the figure the work done during the cycle is A. ZeroB. 2 PVC. 6 PVD. 9 PV

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`Delta S` for the reaction , `MgCO_(3)(s)rarr MgO(s)+CO_(2)(g)` will be :

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The heat of combustion of `H_(2)(g)` at constant pressure and `300k` is `-280kJ mol^(-1)`. What will be the heat of combustion at constant volume and at `300K`?

The standard absolute entropy of a substance, `(S^(Theta))` is the entropy of the substance in its standard state at `1` atm, temperature beingA. `0K`B. `298K`C. `398K`D. `273 K`

Heat of reaction for `C_(6)H_(12)O_(6(s))+6O_(2(g)) rarr 6CO_(2(g))+6H_(2)O_(v)` at constant pressure is `-651 kcal` at `17^(@)C`. Calculate the heat of reaction at constant volume at `17^(@)C`.

The enthalpy change of a reaction does not depand onA. State of reactants and productsB. Nature of reactants and productsC. Different intermediate reactionsD. Initial and final enthalpy change of reaction

What is Equation of state and examples ?

What is meant by Thermodynamic equilibrium ?

Difference between heat and work.

What is Zeroth Law of Thermodynamics ?

Define Thermodynamic variables. Give the state of Thermodynamic variables.

Heat of dissociation of acetic acidis 0.30 kcal`mol^(-1)` . Hence,enthalpy change when 1 mol of`Ca(OH)_(2)` is completely neutralized by acetic acid would beA. `-13.4 kcal`B. `-27.1 kcal`C. `-26.8 kcal`D. `-27.4 kcal`

The enthalpy change when 1 L of 1 M `H _(2)SO_(4)` is completely neutralized by 1L of1 M `Ca(OH)_(2)` will beA. `-13.7 kcal`B. `-27.4 kcal`C. `-1.37 kcal`D. `- 2.74 kcal`

The enthalpies of elements in their standard states are takeen as zero. The enthalpy of formation of a compoundA. is always negativeB. is always positiveC. may be positive or negativeD. is never negative

ΔU of combustion of methane is -x kJ mol-1. The value of ΔH° is(a) = ΔU°(b) &gt; ΔU°(c) &lt; ΔU°(d) = 0

The enthalpies of all elements in their standard states are:(a) unity(b) zero(c) &lt; 0(d) different for each element

For the process to occur under adiabatic conditions, the correct condition is(a) ΔT = 0(b) Δp = 0(c) q = 0(d) W = 0

A thermodynamic state function is a quantity:(a) used to determine heat changes(b) whose value is independent of path(c) used to determine pressure volume work(d) whose value depends on temperature only

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Calculate `Delta H ^(@)` for the reaction `CH_(2)= CH_(2) + 3O_(2) rarr 2CO_(2)+ 2H_(2)O` Given that the average bond energies of the different bonds are `:` `{:("Bond",C-H,O=O,C=O,O-H,C=C),("Bond energy " ( kJ mol^(-1)),414,499,724,460,619):}`

Hydrazine , `N_(2)H_(4) ,` contains aN-N single bond and 4N -H bonds . Use bond energies to calculate `DeltaH`in KJ for the reaction : `N_(2)+2H_(2)toN_(2)H_(4)` `{:(Bond ,"Energies" (KJ.mol^(-1))),(H-H,436),(H-N,386),(N-N,193),(N=N,418),(N-=N,941):}`A. `-425KJ`B. `-76KJ`C. `76KJ`D. `245KJ`

Find the mass of all molecules leaving one square centimetre of water surface per second into a saturated water vapour above it at a temperature `t = 100 ^@C`. It is assumed that `eta = 3.6 %` of all water vapour molecules falling on the water surface are retained in the liquid phase.

ΔU of combustion of methane is -x kJ mol-1. The value of ΔH° is(a) = ΔU°(b) > ΔU°(c) < ΔU°(d) = 0

The enthalpies of all elements in their standard states are:(a) unity(b) zero(c) < 0(d) different for each element