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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1201. |
Statement-1. Enthalpy change at constant pressure is always greater than enthalpy change at constant volume for any reaction. Statement -2. Work is done by the system at constant pressure for a given change in volume but no work is done by the system at constant volume.A. Statement -1 is True, Statement-2 is True, Statement-2 is a correct explanation of Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement -1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - d | |
| 1202. |
Using the data ( al values are in kilocalories per moleat `25^(@)C `) given below, calculate the bond energy of `C-C` and `C-H` bonds. `DeltaH^(@) ` combustion( ethane) `= - 372.0` `DeltaH^(@) ` combustion ( propane) `= - 530.0``DeltaH^(@) ` forC ( graphite` rarr C (g) = 172.0` Bond energy of `H-H= 104.0` `Delta_(f)H^(@)` of`H_(2)O(l)= - 68.0, Delta_(f)H^(@)` for`CO_(2) = - 94.0` |
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Answer» We are giving (i)`C_(2)H_(6)(g)+(7)/(2) O_(2)(g) rarr 2CO_(2)(g)+H_(2)O(l),DeltaH^(@) =- 372.0 kcal` (ii) `C_(3)H_(8) +5O_(2)(g) rarr 3CO_(2)(g)+4H_(2)O(l),DeltaH^(@) = - 530.0 kcal` (iii) `C(s) rarr C(g), DeltaH^(@) =172.0kcal` (iv) `H_(2)(g) rarr 2H(g) , DeltaH^(@) = 104.0kcal` (vi)` C(g)+ O_(2)(g) rarr CO_(2)(g) , DeltaH^(@) = - 94.0 kcal` Suppose the bond energy of`C-C` bond `=x kcal ` mol`e^(-1)` and thatof`C-H` bond `=y` kcal` mol e^(-1)`. Then for `C_(2)H_(6)(g), i.e.,H- underset(H) underset(|) overset(H) overset(|) (C) - underset( H ) underset(|) overset(H) overset(|) (C)-H rarr C(g) + H(g),DeltaH = x+6y` ....(vii) andfor `C_(3)H_(8) (g)`, i.e., `H-underset(H) underset(|) overset(H) overset(|)(C) - underset(H) underset(|) overset(H) overset(|) (C) - underset(H) underset(|) overset(H) overset(|) (C) - H rarr 3C(g) +8 H(g), DeltaH = 2x + 8 y ` .....(viii) To get Eqn. (vii) , operate Eqn. (i) `+ 2 xx `Eqn. (iii) ` + 3 xx ` Eqn. (iv) `- 3 xx` Eqn. (v) `- 2 xx` Eqn. (vi). This gives`DetlaH = 676 kcal` To get Eqn. (viii) , operateEqn. `+` Eqn. (iii) `+`Eqn. (iii) `+ 4 xx` Eqn. (vi)` - 4 xx` Eqn. (v) `- 3 xx `Eqn. (vi). This gives `DeltaH = 956kcal` Thus, `x + 6y= 676 ` `2x + 8y=956` On solving these equations, we get `x= 82,y = 99` Hence, `C-C` bond energy `= 83 kcal mol^(-1)` and `C-H` bond energy `= 99` kcal `mol^(-1)` |
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| 1203. |
At a certain temperature ‘T’ endothermic reaction A → B proceeds virtually to end. Determine the sign of ΔS for the reaction A → B and ΔG for the reverse reaction B →A. |
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Answer» ΔS – +ve ΔH = +ve, ΔG = -ve the reaction proceeds virtually to end in reverse reaction B → A, ΔS = -ve, ΔG = +ve |
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| 1204. |
If `Delta H gt 0` and `Delta S gt 0`, the reaction proceeds spontaneously when :-A. `Delta H gt 0`B. `Delta H lt T Delta S`C. `Delta H = T Delta S`D. None |
| Answer» Correct Answer - B | |
| 1205. |
At a certain temperature T, the endothermic reaction `A rarr B` proceeds almost to completion.The entropy change is :A. `Delta S = 0`B. `Delta S lt 0`C. `Delta S gt 0`D. Cannot be predicted |
| Answer» Correct Answer - C | |
| 1206. |
For the process `H_(2)O(l) (1 "bar", 373 K) rarr H_(2)O(g) (1"bar", 373 K)` the correct set of thermodynamic parameters isA. `DeltaG=0, DeltaS=+ve`B. `DeltaG=0,DeltaS=-ve`C. `DeltaG=+ve,DeltaS=0`D. `DeltaG=-ve,DeltaS=0` |
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Answer» Correct Answer - a `underset((1 "bar", 373K))(H_(2)O_((l)))hArrunderset((1 "bar", 373 K))(H_(2)O_((g)))` At `100^(@)C H_(2)O_((l))` is in equilibrium with `H_(2)O_((g))`,therefore `DeltaG=0` Because liquid molecules are converting into gases molecules therefore `DeltaS=+ve` |
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| 1207. |
In which of the following chemical reactions, `deltaHgtdeltaU?`A. `H_(2)(g)+I_(2)(g)to2HI(g)`B. `N_(2)(g)+3H_(2)(g)to NH_(3)(g)`C. `2H_(2)O_(2)(g) to 2H_(2)O(g)+ O_(2)(g)`D. `C_(s)(g)+O_(2)(g)to CO_(2)(g)` |
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Answer» Correct Answer - C |
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| 1208. |
For a particular reversible reaciton at temperature `T, DeltaH` and `DeltaS` were found to be both `+ve`. If `T_e` is the temperature at equilibrium, the reaciton would be spontaneous when :A. `T_(e)gtT`B. `TgtT_(e)`C. `T_(e)` is `5` times `T`D. `T=T_(e)` |
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Answer» Correct Answer - b `DeltaG=DeltaH-TDeltaS` at equilibrium, `DeltaG=0` so, `TDeltaS=DeltaH` As `DeltaH` and `DeltaS` are `+ve`, for a reaction to be spontaneous `DeltaG` should be -ve. `TDeltaSgtDeltaH i.e,. TDeltaSgtT_(e)DeltaS` `:. TgtT_(e)` |
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| 1209. |
What will be the melting point of KCl if enthalpy change for the reaction is `"7.25 J mol"^(-1)` and entropy change is `"0.007 J K"^(-1)"mol"^(-1)`?A. 1835.2 KB. 173 KC. 1035.7 KD. 1285.2 K |
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Answer» Correct Answer - C `T=(DeltaH)/(DeltaS)=(7.25)/(0.007)= 1035.7 K` |
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| 1210. |
Calculate the entropy change accompanying the following change of state `H_(2)O (s, 10^(@)C, 1 atm) rarr H_(2)O(l, 10^(@)C, 1atm)` `C_(P)` for ice `= 9 cla deg^(-1) mol^(-1)` `C_(P)` for `H_(2)O = 18 cal deg^(-1) mol^(-1)` Latent heat of fustion of ice `= 1440 cal mol^(-1) at 0^(@)C`. |
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Answer» The total process involved three stages and entropy change can be calculated for each stage as follows: `[DeltaS_(1) rArr` For changing `1` mole of ince from `-10^(@)C, 1atm` to `0^(@)C, 1atm]` `DeltaS_(1) = int_(-10)^(0) n.(C_(P))/(T).dT` `=n xx C_(P) xx 2.303 xx "log"(273)/(263)` `= 1 xx 9 xx 2.303 xx 0.0162` `= 0.336 cal deg^(-1) mol^(-1)` `DeltaS_(2) rArr` For melting `1` mol of ice at `0^(@)C` will be as `DelatS_(2) = (q_(rev))/(T) = (1440)/(273) = 5.25 cal deg^(-1) mol^(-1)` `DeltaS_(3) rArr` For heating `1` mol of `H_(2)O` from `0^(@)C` to `10^(@)C` at `1 atm` `DeltaS_(3) = int_(0)^(10) (nC_(P))/(T) dT` `= 1 xx 18 xx 2.303 "log" (283)/(273)` `= 0.647 cal deg^(-1) mol^(-1)` `DeltaS = 0.336 +5.276 + 0.647` `= 6.258 cal deg^(-1) mol^(-1)` |
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| 1211. |
Match the transformation in colums I with appropriate options in column II. `{:("Column I" ,"Column II"),((A)CO_(2)(s) rarr CO_(2)(g),(p)"phase transition"),((B)CaCO_(3)(s)rarr CaO(s) + CO_(2)(g) ,(q)"allotropic change"),((C)2H.rarrH_(2)(g),(r)DeltaH "is positive"),((D)P_(("white, solid"))rarr P_(("red,solid")),(s)DeltaS "is positive"),(,(t) DeltaS "is negative"):}` |
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Answer» Correct Answer - A::B::C::D `(A) CO_(2)(s) rarr CO_(2)(g)` It is phase transition . The process is endothermic (sublimation). Gas is produced , so entropy increases. (B) On heating `CaCO_(2)` decomposes. So process is endothermic . The enthropy increases as gaseous product is formed. (C) `2H.rarr H_(2)(g)` Entropy decreases as number of gaseous particles decreases. (D) It is phase transition. Red P is more stable than while. So `DeltaH "is" -ve`. |
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| 1212. |
Which of the following preocesses are accompained by increase of entropy. a. Dissolution of iodine in a solvent b. `HCI` is added to `AgNO_(3)` and a precipitate of `AgCI` is obtained. c. A partition is removed to allow two gases to mix. |
| Answer» Increase of entropy: (a) and (c ). | |
| 1213. |
The following data is known about the melting of `KCl : Delta h = 7.25 kJ mol^(-1) ` and `Delta S = 0.007 kJ K^(-1) mol^(-1)` . Calculate the melting point. |
| Answer» Correct Answer - `1035.7 K` | |
| 1214. |
The following data is known for melting of `KCI: DeltaS^(Theta) = 0.007 kJ K^(-1) mol^(-1), DeltaH^(Theta) = 7.25 kJ mol^(-1)` Calculate the melting point of `KCI`. |
| Answer» `T = (DeltaH)/(DeltaS) = (7.25)/(0.007) = 1035.7 K` | |
| 1215. |
For a particular reversible reaciton at temperature `T, DeltaH` and `DeltaS` were found to be both `+ve`. If `T_e` is the temperature at equilibrium, the reaciton would be spontaneous when :A. `T_(e) gt T`B. `T gt T_(e)`C. `T_(e)` is 5 times TD. `T=T_(e)` |
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Answer» Correct Answer - B `DeltaG=0` at equilibrium |
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| 1216. |
Following data is known about melting of a compound AB`.DeltaH=9.2kJmol^(-1),DeltaS=0.008kJK^(-1)mol^(-1)` Its melting point isA. 736 KB. 1050 KC. 1150 KD. `1150^(@)C` |
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Answer» Correct Answer - C `DeltaG=DeltaH-T Delta S` |
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| 1217. |
Which of the following conditions leads to a spontaneous process at all temperatureA. `Delta H` is positive and `Delta S` is negativeB. `Delta H` is negative and `Delta S` is positive.C. Both `Delta H` and `Delta S` are negative.D. Both `Delta H` and `Delta S` are positive |
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Answer» Correct Answer - B `Delta G = Delta H = T Delta S` `= (-ve) - (+ ve) (+ ve)` `= - ve` Thus, `Delta G` is always negative, reaction proceeds spontaneously at all temperatures. |
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| 1218. |
Match the following : |
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Answer» Correct Answer - (A-p,q,s,u); (B-s, t, u); (C-r,t); (D-s,t) `Delta_(f)H^(@)` and `Delta_(f)G^(@)` are zero for species in their standard states from III law entropy of a perfectly crystalline solid is zero at 0 K. |
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| 1219. |
For a spontaneous reaction the `DeltaG`, equilibrium constant (K) and `E_("cell")^(@)` will be respectivelyA. `-ve, gt 1, -ve`B. `-ve, lt 1, -ve`C. `+ve, gt 1, -ve`D. `-ve, gt 1, +ve` |
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Answer» Correct Answer - D `DeltaG=G_(P)-G_(R)` |
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| 1220. |
Temperature of one mole of helium gas is increased by `1^(@)C`. Hence,the increasein its internal energy will beA. 2 calB. 3 calC. 4 calD. 5 cal |
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Answer» Correct Answer - B `C_(v) = ((DeltaU)/(DeltaT))_(V)`or` DeltaU = C_(v)DeltaT` But for monoatomic gas, `C_(v)= (3)/(2) R`. Also `DeltaT =1^(@)` (Given) Hence, `DeltaU = C_(v)DeltaT = (3)/(2) R = (3)/(2) xx2 = 3 cal` |
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| 1221. |
What will be the value of `Delta G^(@)`. If equilibrium constant for a reaction is 10.A. `-50.44 kJ mol^(-1)`B. `-5.705 kJ mol^(-1)`C. `-25.44 kJ mol^(-1)`D. `-10 kJ mol^(-1)` |
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Answer» Correct Answer - B `DeltaG=-2.303 RT log K_(C)` |
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| 1222. |
Combustion of hydrogen in a fuel cell at 300 K is represented as `2H_(2(g))+O_(2(g)) rarr 2H_(2)O_((g))`. If `Delta H` and `Delta G` are `-241.60 kJ mol^(-1)` and `-228.40 kJ mol^(-1)` of `H_(2)O`. The value of `DeltaS` for the above process isA. 4.4 kJB. `-88 J`C. `+88 J`D. `-44 J` |
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Answer» Correct Answer - B `DeltaG=DeltaH-T Delta S` |
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| 1223. |
The temperature of `1` mole helium gas is increased by `1^@C`. Find the increase in internal energy. |
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Answer» Correct Answer - 3 `((DeltaE)/(DeltaT))_(V)=C_(V)` `:. Delta E=C_(V) DeltaT` `C_(V)=3/2 R` for monatomic gas `DeltaT= 1K` `:. DeltaE=3/2 R=3/2 xx2=3 cal` |
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| 1224. |
Identify the correct statement regarding a spontaneous process.A. Lowering of energy in the process is the only criterion for spontaneity.B. For a spontaneous process in an isolated system, the change in entropy is positive.C. Endothermic process are never spontaneous.D. Exothermic process are always spontaneous. |
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Answer» Correct Answer - B II Law |
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| 1225. |
The entropy change of orgon is given to a good approximation by the expression : `S_(m)//JK^(-1)mol^(-1)=36+20lnT//K` Calculate the change in Gibbs free energy of one mole of argon gas when it is heated from `300` K to `400` K at constant pressure. Express your answer excluding sing and decimal places. |
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Answer» Correct Answer - 15 |
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| 1226. |
Which fo the following expression defines the physical siginificance of Gibbos energy changeA. `Delta G = W ("exp")`B. `Delta (G) = W ("nonexp")`C. `Delta G = - W ("exp")`D. `- Delta G = W ("nonexp")` |
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Answer» Correct Answer - D A knowledge of the magnitude of the free enegy change of a chemical reaction allows the predicition of the useful work (other than pressure -voume work or expanision work) that can be extracted from it. It can be shown that a decrease in Gibbs energy for a process is equal to the maximum possible useful work that can be derived fromk the process, i.e., `- Delta G = W_(max)` for a reversible change at constant `T` and `P`. In case of galvanic cells (chapter 8), Gibbs energy change, `Delta G`, is related to the electrical worik done by the cell. If `E` is the emf of the cell and `n` moles of electrons are involved, the electrical work will be equal to `nFE` (where `F` is the Faraday constant). Thus. `Delta_(r) G = - nFE_(cell)` If the reactants and products are in their standard states, then `Delta_(r) G^(@) = - nFE_(cell)^(@)` Here, `E_(cell)^(@)` is the standard cell potential. |
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| 1227. |
An ideal gas is taken through the cycle `A rarr B rarr C rarr A` As shown in figure. If net heat supplied to the gas in the cycle is 5j. Find the work done by the gas in the process `C rarr A` in Joule (taken mole value) |
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Answer» Correct Answer - 5 `W_(AB)=P Delta V, W_(BC)=0` `q= DeltaU+(-W_(T))` `:. DeltaU=0` `:. q=-W_(T)=-[W_(AB)+W_(BC)+W_(CA)]` `rArr 5=-[10+0+W_(CA)]` or `W_(CA)=-5j` |
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| 1228. |
Which of the following expression defines the physical significance of Gibbs energy changeA. `-DeltaG=W_("Compression")`B. `DeltaG=W_("expansion")`C. `DeltaG=-W_("expansion")`D. `-DeltaG=W_("non-expansion")` |
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Answer» Correct Answer - D `-DeltaG=W_("max")` |
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| 1229. |
If 2kcal heat is given to a system and 6 kcal work is done on the system then the internal energy of system will increase by how many kcal? |
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Answer» Correct Answer - 8 `= Delta E-W` |
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| 1230. |
For which of these processes is the value of `Delta S` negative? i. Sugar is dissolved in water. ii. Steam condenses on a surface. iii. `CaCO_(3)` is decomposed into `CaO` and `CO_(2)`.A. i onlyB. ii onlyC. ii and iii onlyD. ii and iii only |
| Answer» In condensation water, vapour converts to liquid water. | |
| 1231. |
Calculate `DeltaG^(Theta)` for the following reaction: `CO(g) +((1)/(2))O_(2)(g) rarr CO_(2)(g), DeltaH^(Theta) =- 282.84 kJ` Given, `S_(CO_(2))^(Theta)=213.8 J K^(-1) mol^(-1), S_(CO(g))^(Theta)= 197.9 J K^(-1) mol^(-1), S_(O_(2))^(Theta)=205.0 J K^(-1)mol^(-1)`, |
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Answer» `DeltaS^(Theta) = sum S^(Theta) ("products") -sumS^(Theta) ("reactants")` `=[S^(Theta)underset(CO_(2)).]-[S^(Theta)underset(CO).+(1)/(2)S^(Theta)underset(O_(2)).]` `= 213.8 - [197.9 +(1)/(2)205] =- 86.6 J K^(-1)` According to Gibbs-Helmholtz equation, `DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)` `=- 282.84 - 298 xx (-86.6 xx 10^(-3))` `=- 282.84 +25.87` `=- 257.033 kJ` |
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| 1232. |
The incorrect expression among the following isA. `(DeltaG_("system"))/(DeltaS_("total"))=-T`B. In isothermal process `W_("reversible")=-nRT` in `V_(f)/V_(i)`C. In `K=(DeltaH^(@)-T Delta S^(@))/(RT)`D. `K=e^(-DeltaG^(@)//RT)` |
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Answer» Correct Answer - C `DeltaG^(@)=-RT ln K_((eq))` |
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| 1233. |
Assertion : An exothermic process which is non-spontaneous at high temperature may become spontaneous at low temperature. Reason : Spontaneous process is an irreversible process and may be reversed by some external agency.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B `DeltaG=DeltaH-TDeltaS` For exothermic process, `DeltaH=-ve`. For `DeltaG` to be `-ve` (for spontaneous process), `DeltaH gt T DeltaS` which is possible only at low temperature. |
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| 1234. |
Temperature of 1.5 mole of gas is increased by `10^(@)C` at constant pressure. Magnitude if work involved is:A. `15R`B. `10R`C. `10R`D. `20R` |
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Answer» Correct Answer - A |
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| 1235. |
The `P - V` diagram of 2 gm of helium gas for a certain process `A rarr B` is shown in the figure. What is the heat given to the gas during the process `A rarr B`? A. `4P_(0)V_(0)`B. `6P_(0)V_(0)`C. `4.5P_(0)V_(0)`D. `2P_(0)V_(0)` |
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Answer» Correct Answer - B |
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| 1236. |
Assertion : Heat added to a system at lower temperature causes greater randomness than when the same quantity of heat is added to it at higher temperature. Reason : Entropy is a measure of the degree of randomness or disorder in the system.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B `DeltaS` is related with q and T for a reversiblereaction as: `DeltaS=(q_("rev"))/(T)` |
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| 1237. |
Assertion : In the proces, `H_(2(g)) rarr 2H_((g))`.entropy increases. Reason : Breaking of bonds is an endothermic process.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B In this process, one molecule of hydrogen gives two atoms i.e., number of gaseous particles increases leading to more disordered state. |
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| 1238. |
Ration of `C_(p)//C_(v) "for" NH_(3)` gas(assuming ideal behaviour) when vibrational degree of freedom are active :A. `(10)/(9)`B. `(4)/(3)`C. `(11)/(8)`D. `(12)/(10)` |
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Answer» Correct Answer - A |
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| 1239. |
Calculate entropy charge `H_(2)O(l,1 atm, 100^(@)C)` rarr `H_(2)O(g, 1" "atm, 110^(@))` `H_(2)O(l, 2" "atm, 100^(@))`rarr `H_(2)O(g, 2" "atm, 100^(@))` `DeltaH_(vap)=40kJ//"mole"`" "`C_(p)(l)=75J//"mole"//K`" "`C_(p)(g)=35 J//"mole"//K` |
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Answer» `underset((A))(H_(2)O(l, 2 atm, 100^(@)C))to underset((B)) (H_(2)O(g, 1 atm,100^(@) C)) to underset((C))(H_(2) O(g, 2 atm, 100^(@)C))` `DeltaS_(A to B)=(DeltaH_("vap"))/(T)= (40xx1000)/(373)` `DeltaS_(B to C) n Rln .(P_(2))/(P_(1))= 1 xx R ln .(1)/(2)` |
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| 1240. |
Value of `gamma` for `CH_(4)` molecule is (Consider vibarational degree of freedom to be active).A. `4//3`B. `(13)/(12)`C. `(15)/(14)`D. `(17)/(15)` |
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Answer» Correct Answer - B `C_(V) =3R+9R rArr12 R` `C_(V) = 13 R` `gamma=(C_(P))/(C_(V))=(13)/(12)` |
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| 1241. |
1 mole of an ideal gas is expanded isokthermally and reversibly from 1 litre to litre. Which of the following is false for the process ?A. `DeltaT=0`B. `DeltaE= 0`C. `DeltaH=0`D. heat supplied (q)=0 |
| Answer» Correct Answer - D | |
| 1242. |
One mole of a real gas is subjected to a process from (2 bar , 30 lit ., 300k) to (2 bar , 40 , lit ., 500K) Given :`C_(v)=25 J//"mole"/K` `C_(p)=40 J//"mole"/K` Calculate `DeltaU`.A. 5000 JB. 6000 JC. 8000 JD. 10000 J |
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Answer» Correct Answer - B `DeltaU=q+"w"` `=(40xx200)+(-2xx10xx100)` `= 6000 J ` |
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| 1243. |
State whther each of the following will increase or decreases the total enegry content of the system: a. Heat transferred to the surroundings b. Work done on the system. c. Work done by the system. |
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Answer» a. When heat is transferred to the surroundings, the total energy content of the system decreases. The system loses a part of energy in the form of heat. b. When work is done on the system, the energy content of the system increases. c. When work is done by the system, the energy content of the system decreases. The system does work at the cost of its energy. |
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| 1244. |
Calculate the entropy change for the following reaction `H_(2)(g) +CI_(2)(g) rarr 2HCI (g) at 298 K` Given `S^(Theta)H_(2) = 131 J K^(-1) mol^(-1), S^(Theta)CI_(2) = 233 J K^(-1) mol^(-1)`, and `S^(Theta) HCI = 187 J K^(-1) mol^(-1)` |
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Answer» `DeltaS^(Theta) = sum S^(Theta)(Products) - sum S^(Theta) (Reactant)` `= 2xx187 -(131 +223)` `= 374 - 131 - 223 = 20 J K^(-1) mol^(-1)` |
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| 1245. |
Calculate heat of dissociation for acetic acid from the following data: `CH_(3)COOH +NaOH rarr CH_(3)COONa +H_(2)O DeltaH = - 13.2 kcal` `H^(o+) +overset(Theta)Ohrarr H_(2)O, DeltaH =- 13.7 kcal` Also calculate heat of dissociation for `NH_(4)OH` if `HCI +NH_(4)OH rarr NH_(4)CI +H_(2)O, DeltaH =- 12.27 kcal` |
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Answer» `CH_(3)COOH(w_(A)) +overset(Theta)OH (s_(B)) rarr CH_(3)COO^(Theta) = H_(2)O, DeltaH_(1) =- 13.2 kcal` `H^(o+)(S_(A)) +overset(Theta)OH(S_(B)) rarr H_(2)O DeltaH_(2) = 13.7 kcal` `DeltaH = DeltaH_(1) - DeltaH_(2)` `=- 13.2 - (-23.7) = 0.5 kcal` Similarly, for `NH_(4)OH:` `Na_(4)OH rarr NH_(4)^(o+) +overset(Theta)OH, DeltaH = ?` `H^(o+)(S_(A)) +NH_(4) OH (W_(B)) rarr NH_(4)^(o+) +H_(2)O, DeltaH_(1) = 12.27` `H^(o+)+overset(Theta)OH rarr H_(2)O` `{:ulbar(NH_(4)OH rarr NH_(4)^(o+)+overset(Theta)OH):}` `DeltaH = DeltaH_(1) - DeltaH_(2) = - 12.7 -(-13.7) = 1.43 cal` |
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| 1246. |
Two moles of an ideal monoatomic gas occupy a volume 2V at temperature 300K, it expands to a volume 4V adiabatically, then the final temperature of gas isA. 179 KB. 189 KC. 199 KD. 219 K |
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Answer» Correct Answer - B Since in an adiabatic process `T_(1)V_(1)^(gamma-1)=T_(2)V_(2)^(gamma-1)` `T_(2)=T_(1)(V_(1))/(V_(2))^(gamma-1)=300((2V)/(4V))^((5)/(3)-1))` `=300((1)/(2)^(2//3)=(300)/(2^(2//3))=188.98 = 189 K` |
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| 1247. |
Which one of the following statements is false?A. Work is a state function.B. Temperature is a state funciton.C. Change in the state is completely defined when the initial and final states are specified.D. Work appears at the boundary of the system. |
| Answer» Work is not a state function, but it is a path function. | |
| 1248. |
A gas expands by `0.372` litre against a constant pressure of `1atm`. Find the work done in cal. |
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Answer» Correct Answer - 9 `W=-P_(ext)xxDeltaV` `=-1xx0.372=-0.372` litre-atm `0.372` litre atm`=-0.372xx1.987/0.0821~~-9.0 cal` |
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| 1249. |
The difference between the heats of reaction at constant pressure and a constant volume for the reaction `2C_(6)H_(6)(l) + 15O_(2)(g) rarr 12CO_(2)(g) + 6H_(2)O(l)` at `25^(@)C` in `kJ` isA. `-7.43`B. `+3.72`C. `-3.72`D. `+7.43` |
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Answer» `2C_(6)H_(6)(l) +15O_(2) rarr 12CO_(2)(g) +6H_(2)O(l)` `DeltaH = DeltaE +Deltan_(g)RT` `DeltaH - DeltaE = Deltan_(g)RT` `Deltan_(g) = 12 - 15 =- 3` `DeltaH - DeltaE =- 3 xx 8.314 xx 298 =- 7432.716 J` |
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| 1250. |
q.w, `DeltaE` and `DeltaH` for the following process ABCD on a monoatomic gas are: A. `w= - 2P_(0)V_(0) ln 2`,B. `w= -2 P_(0)V_(0) ln 2`C. `w= P_(0)V_(0) (1+ ln 2)`,D. `w= -P_(0)V_(0)ln2` |
| Answer» Correct Answer - A | |