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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.
| 1951. |
`{:(C(s)+O_(2)(g)rarrCO_(2)(g), ,,,,DeltaH= -94.3 kcal//mol),(CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g),,,,,DeltaH= -67.4 kcal//mol),(O_(2)(g)rarr2O(g),,,,,DeltaH=117.4 kcal//mol),(CO(g)rarrC(g)+O(g),,,,,DeltaH=230.6 kcal//mol):}` Calculate `Delta H` for `C(s)rarrC(g)` in `kcal//mol`.A. 171B. 154C. 117D. 145 |
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Answer» Correct Answer - D `C(s)rarrC(g)" "`can be obtained as, `Delta H=DeltaH_(1)-DeltaH_(2)-(1)/(2)DeltaH_(3)+DeltaH_(4)` |
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| 1952. |
Compute the standard free enegry of the reaction at `27^(@)C` for the combustion fo methane using the give data: `CH_(4)(g) +2O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l)` `{:(Species,CH_(4),O_(2),CO_(2),2H_(2)O(l)),(Delta_(f)H^(Theta)(kJmol^(-1)),-74.8,-,-393.5,-285.8),(S^(Theta)(JK^(-1)mol^(-1)),186,205,214,70):}` |
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Answer» `DeltaH^(Theta) = Delta_(f)H^(Theta)underset((CO_(2))).+2Delta_(f)H^(Theta)underset((H_(2)O)).-Delta_(f)H^(Theta)underset((CH_(4)))` `=- 393.5 +2 xx (-285.8) -(-74.8)` `=- 890 kJ mol^(-1)` `DeltaS^(Theta)=S^(Theta)underset((CO_(2))).+2S^(Theta)underset((H_(2)O)).-S^(Theta)underset((CH_(4))).-2S^(Theta)underset((O_(2))."sup")` `=214 +2 xx 70 - 186 -2xx 205` `=- 242 J K^(-1)mol^(-1)` `DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)` `=-890 - 300 xx (-242 xx 10^(-3))` `=- 890 + 72.6 =- 817.4 kJ mol^(-1)` |
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| 1953. |
A chemical reaction is spontaneous at 298 K but non-spontaneous at 350 K . Which one of the following following s true for this reaction ?A. `{:(Delta G,DeltaH ,DeltaS),(-,-,+):}`B. `{:(Delta G,DeltaH ,DeltaS),(+,+,+):}`C. `{:(Delta G,DeltaH ,DeltaS),(-,+,-):}`D. `{:(Delta G,DeltaH ,DeltaS),(-,-,-):}` |
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Answer» Correct Answer - D As reaction is spontaneous at low temperature and non-spontaneous at high temperature,i.e., `DeltaG` is-ve at low temperature and `+ve` at high temperature and `DeltaG =DeltaH - T DeltaS`, only ( e) in correct. |
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| 1954. |
For the reaction `N_(2)(g) + 3H_(2) rarr 2NH_(3)(g)` `Delta H = - 95.4 kJ and Delta S = -198.3 JK^(-1)` Calculate the temperature at which Gibbs energy change `(Delta G)` is equal to zero. Predict the nature of the reaction at this temperature and above it. |
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Answer» `Delta G = Delta H - T Delta S` `Delta G = 0` `Delta H - T Delta S = 0` or `Delta H = T Delta S` `T = (Delta H)/(Delta S)=(-95.4 xx 1000 J)/(-198.3 JK^(-1))=481 K` At this temperature, the reaction would be in equilibrium and with the increase in temperature the opposing factor `T Delta S` would become more and hence, `Delta G` would become positive and the reaction would become non-spontaneous. The reaction would be spontaneous at the temperature below 481 K. |
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| 1955. |
For the reaction, `N_(2)(g) +3H_(2)(g) rarr 2NH_(3)(g)` `DeltaH =- 95.0 kJ` and `DeltaS = - 19000 J K^(-1)` Calculate the temperature in centigrade at which it will attain equilibrium. |
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Answer» At equilibrium, `DeltaG = 0` `T = (DeltaH)/(DeltaS) = (-95 xx 10^(3))/(-190000) = 0.5K` |
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| 1956. |
Find out the heat evolved in combustion if `112` litres ( at `STP`) of water gas (mixture of eqal volume of `H_(2)(g)` and `CO(g)))`. `{:(H_(2)(g)+1//2O_(2)(g)rarrH_(2)O(g),,,,DeltaH= -241.8 kJ),(CO(g)+1//2O_(2)(g)rarrCO_(2)(g),,,,DeltaH= -283 kJ):}` |
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Answer» Correct Answer - `1312 kJ` Equal volume of `H_(2)(g) & CO_(g)` Total volume `=112 L` So , volume of `CO= "volume of" H_(2)=56 L` Mole of `CO= "Mole of " H_(2)=2.5 "mole"` `{:(,H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(g),,,DeltaH= -241.8 kJ),("For",1"mole",,,DeltaH= -241.8 kJ),("For",2.5"mole",,,DeltaH= -241.8xx2.5),("Similarly"",",CO_(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g),,,DeltaH= -283 kJ),(,,,,Delta H= -283xx2.5 kJ):}` Total Heat evolved `=[-241.8+(-283]2.5= -1312 kJ` |
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| 1957. |
If agas expands through a volume of 100mL against atmospheric pressure, the work done by the gas `= "…................"` joule . |
| Answer» `10.13 J ( 1L atm= 101.3J)` | |
| 1958. |
The ratio of the relative rise in pressure for adiabatic compression to that for isothermal compression is (A) γ (B) 1/γ (C) 1 - γ (D) 1/(1 - γ) |
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Answer» Answer is (A) γ For adiabatic compression, PVγ = Constant. Hence dp = -γpdV/V. For isothermal compression pV = constant, hence dp = pdV/V. |
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| 1959. |
What is the value of isothermal bulk modulus of an of an ideal gas at one atmosphere pressure (A) 1.01 x 1012 Nm-2 (B) 1.01 x 1010 Nm-2 (C) 1.01 x 106 Nm-2 (D) 1.01 x 105 Nm-2 |
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Answer» Answer is (D) 1.01 x 105 Nm-2. Bulk modulus of elasticity is given by V dp/dV. For isothermal process, PV = k, Hence, Vdp/dV = p. |
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| 1960. |
During adiabatic expansion, the increase in volume is associated with :(A) increase in pressure and temperature(B) decrease in pressure and decrease in temperature. (C) increase in pressure and decrease in temperature(D) decrease in pressure and increase in temperature. |
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Answer» Answer is (B) decrease in pressure and decrease in temperature. |
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| 1961. |
The adiabatic bulk elasticity depends on(A) pressure & volume(B) volume and pressure (C) atomicity & pressure (D) none of the above |
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Answer» Answer is (C) atomicity & pressure |
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| 1962. |
The adiabatic Bulk modulus of a perfect gas at pressure is given byA. pB. 2pC. p/2D. `lambdap` |
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Answer» Correct Answer - D |
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| 1963. |
The adiabatic elasticity of hydrogen gas `(gamma=1.4)` at `NTP`A. `1xx10^(5)N//m^(2)`B. `1xx10^(-8)N//m^(2)`C. `1.4xxN//m^(2)`D. `1.4xx10^(5)N//m^(2)` |
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Answer» Correct Answer - D |
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| 1964. |
Consider the cyclic process `R rarr S rarr R` as shown in the fig. You are told that one of the path is adiabatic and the other one isothermal. Which one of the following is (are) true ? A. Process `R rarr S` is isothermalB. Process `S rarr R` is adiabaticC. Process `R rarr S` is adiabaticD. Such a graph is not possible |
| Answer» Correct Answer - D | |
| 1965. |
Calculate the standard enthalpy change ( in kJ `mol^(-1))` for the reaction , `H_(2)(g) + O_(2)(g) rarrH_(2)O_(2)(g)`, given that bond enthalpy of `H-H,O = O, O-H` and `O-O` ( in `kJ mol^(-1))` are respectively 438, 498,464 and 138A. `-1 30`B. `065`C. `+130`D. `- 334` |
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Answer» Correct Answer - A `H -H(g)+O=O(g) rarr H-O-O-H(g)` `Delta_(r)H= BE(H-H) + BE(O=O)-[2BE(O-H) +BE(O-H)]` `=438 +498 - [2 xx 464 +138 ]kJ mol^(-1)` `= 936- ( 928+ 138 )= - 130 kJ mol^(-1)` |
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| 1966. |
The enthalpy of vaporisation of `C Cl_(4)` is 30.5 kJ `mol^(-1)` . Calculate the heat required for the vaporisation of 284 g of `C Cl_(4)` at constant pressure ( Molar mass of `C Cl_(4) =154 g mol^(-1))` |
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Answer» 1 mole of `C Cl_(4) =154 g` Enthalpy of vaporisation of 154 g of`C C l_(4) = 30.5kJ` `:. `Enthalpy of vaporisation of 284 g of `C C l_(4) = ( 30.5)/( 154) xx 284 kJ = 56.25kJ` |
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| 1967. |
The volume of gas is reduced to half from its original volume. The specific heat will beA. reduced to halfB. be doubledC. remain constantD. increase four times |
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Answer» Correct Answer - C |
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| 1968. |
The enthalpy of vaporisation of `C Cl_(4)` is `30.5 kJ mol^(-1)`. Calculate the heat required for the vaporisation of `284 g " of " C Cl_(4)` at constant pressure. (Molar mass of `C Cl_(4) = 154 g mol^(-1)`) |
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Answer» Given that, 1 mol of `C Cl_(4) = 154 g` `Delta_("vap") H " for " 154 g C Cl_(4) = 30.5 kJ` `:. Delta_("vap") H " for " 284 g C Cl_(4) = (30.5 xx 284)/(154) kJ = 56.25 kJ` |
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| 1969. |
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 |
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Answer» Correct Answer - B |
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| 1970. |
An ideal gas is expanded irrversibly against 10 bar pressure from 20 litre to 30 litre . Calculate if processure is isoenthalpic : |
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Answer» Correct Answer - D |
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| 1971. |
Correct relation among the followingA. `Delta G_("system")=- Delta S_("total")`B. `Delta G_("system")=-T Delta S_("total")`C. `Delta G=Delta H+T Delta S`D. `Delta S=(Delta G+Delta H)/(T)` |
| Answer» Correct Answer - B | |
| 1972. |
At the vicinity of absolute zeroA. `Cp+Cv=0`B. `Cp lt Cv`C. `Cp-Cv =0`D. `Cp gt Cv` |
| Answer» Correct Answer - C | |
| 1973. |
Which of the following in incorrectA. When `Delta G lt 0` process is spontaneousB. When `Delta G gt 0` process is non spontaneousC. When `Delta G=0` process is at equilibriumD. When `Delta G gt 0` process is spontaneous |
| Answer» Correct Answer - D | |
| 1974. |
For a spontaneous process.A. `Delta G_("system")=`+ve onlyB. `Delta G_("system")=` zeroC. `Delta S_("total")=` -veD. `Delta S_("total")=` +ve |
| Answer» Correct Answer - D | |
| 1975. |
For a reaction to occur spontaneouslyA. `(Delta H-T Delta S)` must be negativeB. `(Delta H+T Delta S)` must be negativeC. `Delta H` must be negativeD. `Delta S` must be negative |
| Answer» Correct Answer - A | |
| 1976. |
Write the mathematical relationship between heat internal energy and work done by the system. |
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Answer» The mathematical relationship between heat, internal energy and work done by the system is ΔE = q + w. |
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| 1977. |
Out of carbon (diamond) and carbon (graphite) whose enthalpy of formation is taken as zero and why? |
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Answer» The enthalpy of formation of graphite is taken as zero because it is a more commonly found stable form of carbon. |
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| 1978. |
From thermodynamic point of view, to which system the animals and plant belong? |
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Answer» The animals and plant belong to open system. |
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| 1979. |
Explain the term Bond enthalpy. |
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Answer» It is amount of energy released when gaseous atoms combines to form one mole of bonds between them or heat absorbed when one mole of bonds between them are broken to give free gaseous atoms. Further ∆rH = B.E. (Reactants) - B.E. (Products) |
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| 1980. |
From thermodynamic point to which system the animals and plants belong? |
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Answer» Open system. |
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| 1981. |
The molar entropy content of `1 ` mole of oxygen `(O_(2))` gas at `300 K` and `1 atm` is `250 J mol e^(-1)K^(-1)`. Calculate `DeltaG` when 1 mole of oxygen is expanded reversibility and isothermally from `300K`, 1 atm to double its volume ( Take `R=8.314J mol e^(-1)K^(-1),log e=2.303)`A. 1.728 KJ `"mole"^(-1)K^(-1)`B. 0C. `-1.728 KJ "mole"^(-1)K^(-1)`D. `0.75 KJ "mole"^(-1)K_(1)` |
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Answer» Correct Answer - A `DeltaG=DeltaH - Delta(TS)` `=DeltaH-TDeltaS " "`(isothermal) `0=T DeltaS=-T(int (dq_("rev"))/(T))` `-int dq_("rev")=- q_("rev")=W_("rev")` as process is isothermal so `DeltaE=0=q_("rev")+ W_("rev")` sp `" " DeltaG=- nRT "In" ((V_(f))/(V_(l))) =- RT "In" 2 k=- 8.314 xx300 xx0.693 xx 10^(-3) KJ mol^(-1) K^(-1)` `=1.728 KJ mol^(-1) K^(-1)` |
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| 1982. |
the value of `DeltaG` for the process `H_(2)O(s) rarr H_(2)O` at `1` atm and `260` K isA. `lt 0`B. `= 0`C. ` gt 0`D. Unpredictable |
| Answer» Correct Answer - C | |
| 1983. |
What is a thermodynamic state function? |
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Answer» A function whose value is independent of path. e.g., P, V, E, H |
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| 1984. |
Which of the following processes is a non-spontaneous process?A. Dissolution of salt of sugar in waterB. Mixing of different gases through diffusionC. Precipitation of copper when zinc rod is dipped in aqueous solution of copper sulphateD. Flow of heat from a cold body to a hot body in contact |
| Answer» Correct Answer - D | |
| 1985. |
Define the terms Spontaneous & Non Spontaneous Processes. |
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Answer» A process which can take place by itself is called spontaneous process. A process which can neither take place by itself or by initiation is called non Spontaneous. |
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| 1986. |
State First Law of thermodynamics. |
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Answer» Energy can neither be created nor destroyed. The energy of an isolated system is constant. ∆U = q + w. |
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| 1987. |
When a bottle of perfume is opened, odorous molecules mix with air and slowly diffuse throughout the entire room. Which is not correct for this process?A. `DeltaG=-ve`B. `DeltaGH~~0`C. `DeltaS=-ve`D. `DeltaS=+ve` |
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Answer» Correct Answer - C `DeltaS` increases and thus `+ve`. |
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| 1988. |
The enthalpy of formation of hypothetical `MgCI` is `-125kJ mol^(-1)` and for `MgCI_(2)` is `-642 kJ mol^(-1)`. What is the enthalpy of the disproportionation of `MgCI`.A. `392 kJ mol^(-1)`B. `-392 kJ mol^(-1)`C. `-767 kJ mol^(-1)`D. `-517 kJ mol^(-1)` |
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Answer» `{:(i.,Mg(s)+CI_(2)(g)rarrMgCI_(2)(s),,DeltaH_(1)=-642),(ii.,Mg(s)+(1)/(2)CI(g)rarrMgCI(s),,DeltaH_(2)=-125),(iii.,2MgCIrarrMgCI_(2)+Mg,,DeltaH =?):}` `DeltaH = DeltaH_(1) - 2DeltaH_(2)` `=- 642 -2 xx 2 xx (-125) =- 392 kJ mol^(-1)` |
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| 1989. |
The Gibbs energy change and standard Gibbs energy change for a reaction are same if the reaction quotient `Q` has value equal to :A. `gt1`B. `lt1`C. `0`D. `1` |
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Answer» Correct Answer - D `DeltaG=DeltaG^(@)+RTln Q` |
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| 1990. |
The product of combustion of an aliphatic thiol `(RSH)` at `298K` are `:`A. `CO_(2)(g), H_(2)(g)`, and `SO_(2)(g)`B. `CO_(2)(g), H_(2)O(l)`, and `SO_(2)(g)`C. `CO_(2)(l), H_(2)O(l)`, and `SO_(2)(g)`D. `CO_(2)(g),H_(2)O(l)`, and `SO_(2)(l)` |
| Answer» `C_(2)H_(5)SH +(9)/(2)O_(2)(g) rarr 2CO_(2)(g)+3H_(2)O(l)+SO_(2)(g)` | |
| 1991. |
Thermodynamic parameters for : `HX + H_2O to H_3O^+ + X^-` are given for `HF, HCL, HBr` and `HI`, which acid `(s)` will show the forward reaction? `{:(,HF,HCl,HBr,HI),(DeltaG^(@),+16,-46,-59,-61),(DeltaH^(@),-13,-59,-63,-57),(TDeltaS^(@),-29,-13,-4,+4):}`A. `HF`B. `HCl`C. `HBr`D. `HI` |
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Answer» Correct Answer - A::B::C::D All are acid, thus reaction will go in forward direction. However maximum for `HI` as their `K_(a)` values are `10^(-3),10^(8),10^(10)` and `10^(17)` for `HF,HCl,HBr` and `HI` respectively. |
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| 1992. |
`400 c c` volume of gas having `gamma=5/2` is suddenly compressed to `100 c c`. If the initial pressure is `P`, the final pressure will beA. `P//32`B. `8P`C. `32P`D. `16P` |
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Answer» Correct Answer - C Here, `V_(1)=400 c c, V_(2)= 100 c c, gamma=5/2, P_(1)=P` `P_(2)=?` for adiabatic change, `P_(2)V_(2)^(gamma)=P_(1)V_(1)^(gamma)` `:. P_(2)=P_(1)((V_(1))/(V_(2)))^(gamma)` `P_(2)=P ((400)/(100))^(5//2)` `= Pxx4xx4xxsqrt(4)=32P` |
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| 1993. |
One mole of an ideal gas is heated from 273K to 546K at constant pressure of 1 atmosphere. Calculate the work done by the gas in the process. |
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Answer» Correct Answer - `2.26xx10^(3)J` For one mole of an ideal gas, `(V_(2))/(T_(2))=(V_(1))/(T_(1)) :. V_(2)= (V_(1))/(T_(1))xxT_(2)= 22.4xx10^(-3)xx(546)/(273)` `V_(2)= 44.8xx10^(-3)m^(3)` `W=P(V_(2)-V_(1))` `=1.01xx10^(5)(44.8-22.4)xx10^(-3)` `W=2.26xx10^(3)J` |
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| 1994. |
The bond energies of `H--H` , `Br--Br` and `H--Br` are `433,, 192` and `364KJmol^(-1)` respectively. The `DeltaH^(@)` for the reaction `H_(2)(g)+Br_(2)(g)rarr2HBr(g)` isA. `-261 KJ`B. `+103 KJ`C. `+261 KJ`D. `-103 KJ` |
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Answer» Correct Answer - D For reaction , `H_(2)(g) + Br_(2)(g) to 2HBr(g) DeltaH^(@)=?` `DeltaH^(@) =- [(2xx` bond energy of HBr )-(bond energy of `H_(2) + ` bond energy of `Cl_(2))]` `DeltaH^(@)=-[2xx(364)-(433+ 192)]KJ` `=-[728-(625)] KJ =-103 KJ` |
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| 1995. |
Express the change in internal energy of a system when (i) No heat is absorbed by the system from the surrounding, but work (w) is done on the system. What type of wall does the system have? (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surrounding. What type of wall does the system have? (iii) w amount of work is done by the system and q amount of heat is supplied to the system. What type of system would it be ? |
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Answer» (i) Here, `q = 0 :. Delta U = q+w= 0 + w_(ad) = w_(ad)` As no heat is absorbed by the system, the wall id adiabatic. (ii) Here, ` w= 0 , q= -q :. Delta U = q + w= -q+0= -q` As heat is taken out, the system must be having thermally conducting walls. (iii) `w = -w , q= +q` `:. Delta u = q+ w = q-w` As work is done by the system on absorbing heat, it must be a closed system. |
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| 1996. |
Calculate the internal energy change in each of the following cases `:-` (i) A system absorbs 15 kJ of head and does 5 kJ of work. (ii) 5 kJ of works is done of the system and 15 kJ of heat is given out by the system. |
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Answer» (i) Here, `q= + 15 kJ , w= - 5 kJ` `:.` According to first law of thermodynamics., `Delta U = q+w= 15 + (-5) = 10 kJ` Thus, internal energy of the system increasesby `10 kJ` (ii) Here,` x= + 5 kJ, q = - 15 kJ` `:. `According to first law of thermodynamics, `Delta U = q+ w= -15 + (+5) = - 10kJ` |
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| 1997. |
A chemist while studying the properties of gaseous `C C l_(2)F_(2)`, a chlorofluorocarbon refrigerant, cooled a 1.25 g sample at constant atmospheric prssure of 1.0 atm from 320 K to 293 K . During cooling the sample volume decreased from 274 to 248 mL. Calculate `Delta H ` and `DeltaU` for the chlorofluorocaron for this process. For `C Cl_(2)F_(2), C_(p) = 80.7 J // ( mol K )`. |
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Answer» `DeltaH = q_(p) ` and `C_(p)` is heat evolved or absorbed per mole for `1^(@) ` fall or rise in temperature . Here, fall in temperature `= 320- 293 = 27 K` Molar mass of `C C l_(2)F_(2) = 12 + 2 xx 35.5 + 2 xx 19 = 121 gmol^(-1)` `:. `Heat evolved from 1.25 g of the sample on being cooled from 320K to 293 K at constant pressure`= ( 80.7) /( 121) xx 1.25 xx 27 J = 22.51J` Further, `Delta H = Delta U + P Delta V = 22.51J` `P Delta V = 1 atm xx (( 248 0 274))/(100) L = - 0.026 L atm = - 0.026 xx 101.325 J = - 2.63 J ` `:. - 22.51 = Delta U - 2.63 J` or ` Delta U = - 22.51 + 2.63 J = - 19.88 J ` |
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| 1998. |
Which of the following expressions is not acceptable? (i) `Delta P = P_(f) - P_(i)` (ii) `Delta_(w) = w_(f) - w_(i)` (iii) `Delta q = q_(f) - q_(i)` (iv) `Delta U = U_(f) - U_(i)`A. (i),(ii)B. (ii),(iii)C. (iii),(iv)D. (i),(iv) |
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Answer» Correct Answer - B We cannot write `Delta w = w_(f) - w_(i)` because work is not a state function. Thus, work done not only depends on the initial state and final state, but also depends on how we carry out the process. Similarly, we cannot write `Delta q = q_(g) - q_(i)` because heat is not a state function. Like work, the heat associated with a given process depends on how process is carried out. |
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| 1999. |
The bond energies of `H--H` , `Br--Br` and `H--Br` are `433,, 192` and `364KJmol^(-1)` respectively. The `DeltaH^(@)` for the reaction `H_(2)(g)+Br_(2)(g)rarr2HBr(g)` isA. `- 261 kJ`B. `+ 103 kJ`C. `+ 261 kJ`D. `- 103 kJ` |
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Answer» Correct Answer - D If the reaction is carried out in the gas phase only then the reaction enthalpy `(Delta_(r) H^(@))` can be calculated by the following relationship: `Delta_(r) H^(@) = sum "Bond enthalpies"_("reactants") - sum "Bond enthalpies"_("product")` ` = [Delta_(H - H) H^(@) + Delta_(Br - BR) H^(@)] - (2 Delta_(H - Br)H^(@))` `(433 + 192) - (2) (364)` `= (625) - (728) = - 103 kJ mol^(-1)` |
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| 2000. |
Which one of the ions in the below would have the largest value of enthalpy of hydration ?A. `{:("Ionic radius in nm " ,"Charge of ion"),(0.0065,+2):}`B. `{:("Ionic radius in nm " ,"Charge of ion"),(0.095,+1):}`C. `{:("Ionic radius in nm " ,"Charge of ion"),(0.135,+2):}`D. `{:("Ionic radius in nm " ,"Charge of ion"),(0.169,+1):}` |
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Answer» Correct Answer - A Smaller the size of the ion andgreater the charge on the ion,greater is the enthalpy of hydration. |
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