InterviewSolution
Saved Bookmarks
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.
| 851. |
What is the equilibrium expression for the reaction `P_(4(s)) + 5O_(2(g))hArrP_(4)O_(10(s))` ?A. `K_(c) = (1)/([O_(2)]^(5))`B. `K_(c) = ([P_(4)O_(10)])/(5[P_(4)][O_(2)])`C. `K_(c) = [O_(2)]^(5)`D. `K_(c) = ([P_(4)O_(10)])/([P_(4)][O_(2)]^(5))` |
|
Answer» Correct Answer - D `K_(p) = K_(c)xx(RT)^(Delta n)` `.: delta n = 1 - 2 = - 1` `:. K_(p) = K_(c)xx(RT)^(-1)` Or `(K_(p))/(K_(c)) = (1)/(RT)` |
|
| 852. |
What is the equilibrium expression for the reaction `P_(4(s)) + 5O_(2(g))hArrP_(4)O_(10(s))` ?A. `K_C=([P_4O_10])/([P_4][O_2]^5)`B. `K_C=1/([O_2])^5`C. `K_C=[O_2]^5`D. `K_C=([P_4O_10])/(5[P_4][O_2])` |
|
Answer» Correct Answer - B |
|
| 853. |
What is the equilibrium expression for the reaction `P_(4(s)) + 5O_(2(g))hArrP_(4)O_(10(s))` ?A. `K_(c)=(1)/([O_(2)]^(5)`B. `K_(c)=([P_(4)O_(10)])/(5[P_(4)][O_(2)])`C. `K_(c)=[O_(2)]^(5)`D. `K_(c)=([P_(4)O_(10)])/([P_(4)][O_(2)]^(5))` |
| Answer» Concentration of solid species are supposed to be unity. | |
| 854. |
Consider the following equilibria: `I. A(s) hArr B(s), II. A(s) hArr B(l)` `III. A(l) hArr B(l), IV. A(g) hArr B(g)` Which of the above will be disturbed by an increase in pressure?A. IIB. I, IIC. I, II, III,D. None of these |
|
Answer» Correct Answer - D Because in all the above processes, the reactants and products are in solid and liquid state which are incompressible and has no effect of pressure. |
|
| 855. |
Decomposition of ammonium chloride is an endothermic reaction. The equilibrium may be represented as: `NH_(4)Cl(s) hArr NH_(3)(g)+HCl(g)` A `6.250 g` sample of `NH_(4)Cl` os placed in an evaculated `4.0 L` container at `27^(@)C`. After equilibrium the total pressure inside the container is `0.820` bar and some solid remains in the container. Answer the followings The extent of decomposition can be increased byA. Increasing the temperatureB. Decreasing the temperatureC. Adding more `NH_(4)Cl`D. Removing HCl(g) |
|
Answer» Correct Answer - A Since, the decomposition is an endothermic process hence on increasing temperature the equilibrium shift in forward direction. |
|
| 856. |
Decomposition of ammonium chloride is an endothermic reaction. The equilibrium may be represented as: `NH_(4)Cl(s) hArr NH_(3)(g)+HCl(g)` A `6.250 g` sample of `NH_(4)Cl` os placed in an evaculated `4.0 L` container at `27^(@)C`. After equilibrium the total pressure inside the container is `0.820` bar and some solid remains in the container. Answer the followings The amount of solid `NH_(4)Cl` left behind in the container at equilibrium isA. `2.856`B. `28.56`C. `0.2856`D. `1.320` |
|
Answer» Correct Answer - A moles of gases, `n=(PV)/(RT)=(0.820xx4)/(0.083xx300)=0.132` mol moles of `NH_(3)`= moles of `HCl=0.132/2=0.066` moles of `NH_(4)Cl` decomposed =moles of `NH_(3)=0.066` moles of `NH_(4)Cl` initially present `=6.250/51=0.122` mole of `NH_(4)Cl` left `=0.122-0.066=0.056` Mass of `NH_(4)Cl` left behind `=0.056xx51=2.856 g` |
|
| 857. |
For the reaction `CO(g)+CI_(2)(g)hArrCOCI_(2)(g)` `K_(p)//K_(c )` is equal toA. `1//RT`B. `1.0`C. `sqrtRT`D. `RT` |
|
Answer» Correct Answer - A `CO(g)+CI_(2)hArrCOCI_(2)(g)` `Deltan=1-2=-1,K_(P)=K_(C)(RT)^(Deltan) therefore (K_(P))/(K_(C))=(RT)^(-1)=(1)/(RT)` |
|
| 858. |
For the reaction `PCI_(3)(g) + CI_(2)(g)hArrPCI_(5)(g)`, the value of `K_(p)` at `250^(@)`C is `0.61` `atm^(-1)`. The value of `K_(c)` at this temperature will beA. 15 `(mol//I)^(-1)`B. 26 `(mol//I)^(-1)`C. 35 `(mol//I)^(-1)`D. 52 `(mol//I)^(-1)` |
|
Answer» Correct Answer - B `PCI_(3)(g) + CI_(2)(g)hArrPCI_(5)(g)` `Delta n = -1, K_(p) = 0.61 atm^(-1)` `K_(C) = K_(p)(RT)^(-Delta n)` `= 0.61 (0.0821xx523)^(1) = 26 mol//L` |
|
| 859. |
The gas phase reaction shown is endothermic as written. Which, change(s) will increase the quantity of `CH_(3)CH=CH_(2)` at equilibrium? (P) Increasing the temperature (Q) Increasing the pressure `H_(3)C-CH=CH_(2)hArr` A. P onlyB. Q onlyC. Both P and QD. Neither P nor Q |
|
Answer» Correct Answer - D |
|
| 860. |
An example of a reversible reaction isA. `Pb(NO_(3))_(2)(aq)+2NaI(aq) rarr PbI_(2)(s)+2NaNO_(3)(aq)`B. `AgNO(3)(aq)+HCl(aq) rarr AgCl(s)+HNO_(3)(aq)`C. `2Na(s)+H_(2)O(l) rarr 2NaOH(aq)+H_(2)(g)`D. `KNO_(3)(aq)+NaCl(aq) rarr Kcl(aq)+NaNO_(3)(aq)` |
|
Answer» Correct Answer - D The reactions (a) and (b) precipitation reactions, they are not reversible. Reaction (d) is an -exchange reaction. Hence, it is a reversible reaction. `KNO_(3)(aq)+NaCl(aq) rarr KCl(aq)+NaNO_(3)(aq)` |
|
| 861. |
For a reaction `A(g) hArr B(g)+C(g)`. `K_(p)` at `400^(@)C` is `1.5xx10^(-4)` and `K_(p)` at `600^(@)C` is `6xx10^(-3)`. Which statement is incorrect?A. The reaction is exothermicB. Increase in temperature increases the formation of BC. Increase in pressure increases the formation of AD. Decrease in temperature and increase in pressure shift the equilibrium towards left |
|
Answer» Correct Answer - A `A(g) hArr B(g)+C(g)` a. With increase of temperature `K_(p)` increases, i.e., with increase of temperature, the reaction is favoured in forward direction and hence reaction is endothermic. Thus statement (a) is incorrect. b. Increase of temperature favours forward reaction and hence the formation of B increases. Correct statement. c. `Deltan=1+1-1=1` `Deltan=+ve`, i.e., with the increase of pressure, reaction is favoured backward direction and hence the formation of A(g) increases. Correct statement. d. As from the above statement in (i) and (iii), the reaction is favoured backward with decrease of temperature and increase of pressure. Correct statement. |
|
| 862. |
`K_(p)//K_(c)` for the reaction `CO(g)+1/2 O_(2)(g) hArr CO_(2)(g)` isA. `RT`B. `(RT)^(-1)`C. `(RT)^(-1//2)`D. `(RT)^(1//2)` |
| Answer» `K_(p)//K_(c)=(RT)^(Deltan)` `(Deltan=-(1)/(2))` | |
| 863. |
`K_(p)//K_(c)` for the reaction `CO(g)+1/2 O_(2)(g) hArr CO_(2)(g)` isA. `RT`B. `(RT)^(1//2)`C. `(1)/((RT)^(3))`D. `(1)/sqrt(RT)` |
|
Answer» Correct Answer - D `K_(p)=K_(c )(RT)^(Deltan)` `Deltan=1-(1+1/2)=1/2` `:. K_(p)/K_(c )=1/sqrt(RT)` |
|
| 864. |
For which reaction at equilibrium does a decrease in volume of the container cause a decrease in product(s), quantity at comstant temperature?A. `CaCO_(3)(s)rarrCaO(s)+CO_(2)(g)`B. `2SO_(2)(g)+O_(2)rarr2SO_(3)(g)`C. `HCl(g)+H_(2)O(l)rarrH_(3)O^(+)(aq)+Cl^(-)(aq)`D. `SO_(2)(g)+NO_(2)(g)rarrSO_(3)(g)+NO(g)` |
|
Answer» Correct Answer - A |
|
| 865. |
`2NO(g)+O_(2)(g)hArr2NO_(2)(g)` At a certain temperature the equilibrium concentration for this system are : [NO]=0.25M, `[O_(2)]`=0.24M,`[NO_(2)]`=0.18M. What is the value of `K_(c)` at this temperature?A. 0.063B. `0.50`C. 1.4D. `2.0` |
|
Answer» Correct Answer - B |
|
| 866. |
`K_(p)//K_(c)` for the reaction `CO(g)+1/2 O_(2)(g) hArr CO_(2)(g)` isA. RTB. `(RT)^(-1)`C. `(RT)^(-1//2)`D. `(RT)^(1//2)` |
|
Answer» Correct Answer - C `K_(p)//K_(c) = (RT)^(Delta n), Delta n = - (1)/(2)` |
|
| 867. |
Some quantity of water is contained in a container as shown in figure. As neon is added to this system at constant pressure, the amount of liquid water in the vessel A. increasesB. decreasesC. remains sameD. changes unpredictably |
|
Answer» Correct Answer - B |
|
| 868. |
For the equilibrium `CuSO_(4)xx5H_(2)O(s)hArrCuSO_(4)xx3H_(2)O(s) + 2H_(2)O(g)``K_(p) = 2.25xx10^(-4)atm^(2)` and vapour pressure of water is `22.8` torr at 298 K. `CuSO_(4)` . `5H_(2)O(s)` is efflorescent (i.e., losses water) when relative humidity is :A. less than `63.3%`B. less than `50%`C. less than `66.6%`D. above `66.6%` |
|
Answer» Correct Answer - B `CuSO_(4)` . `5H_(2)O(s)hArrCuSO_(4)` . `3H_(2)O(s) + 2H_(2)O(g)` `K_(p) =2.25xx10^(-4)` `K_(p) = P^(2)H_(2)O =2.25xx10^(-4)` `pH_(2)O = 1.5xx10^(-2)` Vapour pressure `=(22.8)/(760) = 3xx10^(-2)` R.H. `=(P_(H_(2))O)/(V.P.)xx100=50%` Therefore, (b) option is correct. |
|
| 869. |
For the equilibrium `CuSO_(4)xx5H_(2)O(s)hArrCuSO_(4)xx3H_(2)O(s) + 2H_(2)O(g)``K_(p) = 2.25xx10^(-4)atm^(2)` and vapour pressure of water is `22.8` torr at 298 K. `CuSO_(4)` . `5H_(2)O(s)` is efflorescent (i.e., losses water) when relative humidity is :A. less than 33.3%B. less than 50%C. less than 66.6%D. above 66.6 % |
|
Answer» Correct Answer - B |
|
| 870. |
Match the column for the following reaction started with `NH_2COONH_4(s) hArr 2NH_3(g)+CO_2(g)` At equilibrium , partial pressure of `NH_3` and `CO_2` are `P_(NH_3)` and `P_(CO_2)` respectively then : |
|
Answer» Correct Answer - A::B::C::D |
|
| 871. |
The dissociation pressure of silver oxide at `445^(@)C "is" 207` atm. Calculate `DeltaG^(@)` for the formation of `1` mole `Ag_(2)O(S)` from metal and oxygen at this temperature. (`"log" 207=2.315`) |
|
Answer» Correct Answer - `3.8Kcal` `T=445^(@)C=445+273=718K` `P=207`atm `n=1` mole `Ag_(2)OhArr2Ag(s)+(1)/(2)O_(2)(g)` `K_(p)=sqrtPo_(2)=(207)^(1//2)=14.39` `DeltaG^(@)=DeltaG^(@)+2.303RT"log"K_(eq.)` But at eq. `DeltaG=0` `DeltaG^(@)=-2.303RT"log" K_(eq.)` But for formation `2Ag(S)+(1)/(2)O_(2)(g)hArrAg_(2)O(S)` `1` mole `K_(P)=(1)/((Po_(2))^(1//2))=(1)/((207)^(1//2)` `DeltaG^(@)=-(2.303xx8.312xx718)/(2)=xx"log"207` `DeltaG^(@)=6872.17log 207` `DeltaG^(@)=15915.75J` `DeltaG^(@)=3789.46 Cal` `DeltaG^(@)=3.789 KCal` `DeltaG^(@)=3.8KCal` |
|
| 872. |
For equilibrium `ZnSO_(4).7H_(2)O(s)hArrZnSO_(4).2H_(2)O(s)+5H_(2)O(g) K_(P)=56.25xx10^(-10) atm^(5)` and vapour pressure of water is `22.8` torr at `298K.ZnSO_(4).7H_(2)O(s)` is efflorescent (lose water) when relative humidity is `[5sqrt56.25=2.23]`A. less than `70.620%`B. less then `74.60%`C. Above than `74.60%`D. Above 70.60%` |
|
Answer» Correct Answer - B `K_(P)=P_(H_(2)O)^(5)=56.25xx10^(-10)` `P_(H_(2)O)=(56.25xx10^(-10))^(1//5)` `=(56.25)^(1//5)xx10^(-2)` `=2.23xx10^(-2)xx760` `=17.01` torr `% "Relative humidity"=("Partial pressure")/("Vapour pressure")xx100` `=(17.01)/(22.8)xx100=74.60%` (less than `74.60%`) |
|
| 873. |
For the equilibrium : `NH_2COONH_4(s) hArrN_2(g)+3H_2(g)+CO(g)+1/2O_2(g)` the value of `K_(p)` is `27xx2^(lambda//2) "atm"^(11//2)` at 800 K and the equilibrium pressure of 22 atm. The value of `lambda` is :A. 11B. 21C. `5.5`D. `10.5` |
|
Answer» Correct Answer - B |
|
| 874. |
`Na_(2)SO_(4).10H_(2)O(s)hArrNa_(2)SO_(4).5H_(2)O(g) K_(P)=2.43xx10^(-10)` atm^(5) incorrect statement is-A. If partial pressure of `H_(2)O` in container is `3.5xx10^(-2)` atm amount of `Na_(2)SO_(4).5H_(2)O` decreases.B. If `P_(H_(2))O=2.5xx10^(-2)` atm then amount of `Na_(2)SO_(4).5H_(2)O` should increases.C. If `P_(H_(2))O=3xx10^(-2)` atm then both slid does not get altered.D. If `Na_(2)SO_(4).5H_(2)O` is completely removed then partial pressure of `H_(2)O` increases at equilibrium. |
| Answer» Correct Answer - D | |
| 875. |
Equilibrium constant for the following equilibrium is given at `)^(@)C` . `Na_(2)HPO_(4)` . `12H_(2)O(s) hArrNa_(2)HPO_(4)` . `7H_(2)O(s) + 5H_(2)O(g)` `K_(p) = 31.25xx10^(-13)`. At equilibrium what will be partial pressure of water vapour: |
|
Answer» Correct Answer - `5xx10^(-3)` atm. `Na_(2)HPO_(4).12H_(2)O(S)hArrNa_(2)HPO_(4).7H_(2)O(S)+5H_(2)O(g) K_(P)=31.25xx10^(-13)` `K_(P)=(P_(H_(2)O))^(5)` `=(P_(H_(2)O))^(5)=31.25xx10^(-13)` `=(P_(H_(2)O))=(3125)^(1//5xx(10^(-10))^(1//5)` `(P_(H_(2)O))=5xx10^(-3)` |
|
| 876. |
The equilibrium constant `K_(c)` for the reaction, `2NaHCO_(3)(s)hArrNa_(2)CO_(3)(s)+CO_(2)(g)+H_(2)O(g)`A. `K_(c)=([Na_(2)CO_(3)][CO_(2)][H_(2)O])/([NaHCO_(3)]^(2))`B. `K_(c)=([Na_(2)CO_(3)])/([NaHCO_(3)]^(2))`C. `K_(c)=[CO_(2)][H_(2)O]`D. `K_(c)=P_(CO_(2)x)xxP_(H_(2)O` |
| Answer» Correct Answer - A | |
| 877. |
Sufficient amount of a solid `X` is taken in a rigid vessel at `T^(@)C` where it attained the equilibrium: `X(s)hArrY_((g))+2Z(g)` Total pressure was measured. Now the vessel is evcuated and filled with sufficient amount of another solid `V` under same conditions where it attained theequilibrium: Total pressure measured now is found to be double that of previous value. Now, if both `X "&" V` solids are allowed to attain their respective equilibrium together in the same vessel at same temperature, select the correct staement(s):A. `K_(P)` for decompostion reaction of `V(s)=8xxK_(P)` for decomposition reaction of `X (s)`.B. In the`3^(rd)` case (when both solids are simultaneously establishing their equilibrium), `P_(Y)=(1)/(8)P_(W)`.C. `P_(Y) "in case"=(1)/(3sqrt3)xxP_(Y) "in" 1^(st)` case.D. In the `3^(rd)` case,`P_(W),P_(Z)=4.9` |
| Answer» Correct Answer - A::B::C::D | |
| 878. |
How many of water are in vapour phase present inside the vessel containing `1L` water after sufficient time? (Vapour pressure of water at `27^(@)C=3000Pa,R=(25)/(3)J//mol-K`)A. `5xx10^(-4)`B. `120`C. `1.2xx10^(-3)`D. None of these |
|
Answer» Correct Answer - A `n=(PV)/(RT)=(3000xx10^(-3))/((25)/(3)xx300)=1.2xx10^(-3)` moles. |
|
| 879. |
Which oxide of nitrogen is the most stabel?A. `2NO_(2(g))hArrN_(2(g))+2O_(2(g))` , `K=6.7xx10^(16) mol litre^(-1)`B. `2NO_((g))hArrN_(2(g))+O_(2(g))` , `K=2.2xx10^(30) mol litre^(-1)`C. `2N_(2)O_(5(g))hArr2N_(2(g))+5O_(2(g))` , `K=1.2xx10^(34) mol^(5) litre^(-5)`D. `2N_(2)O_((g))hArr2N_(2(g))+O_(2(g))` , `K=3.5xx10^(33) mol litre^(-1)` |
| Answer» Greater is the value of `K`, more will be the tendency to show forward reaction. | |
| 880. |
If the equilibrium constant for the reaction `0.125`. `P_(4(g))+6Cl_(2(g))hArr4PCl_(3(g))` The value of equilibrium for this reactionA. `0.25`B. `8`C. `0.125`D. `6` |
| Answer» Correct Answer - B | |
| 881. |
Consider the equilibrium : `P_((g))+2Q_((g))hArrR_((g))`. When the reaction is carried out at a certain temperature, the equilibrium concentration of `P` and `Q` are `3M` and `4M` respectively. When the volume of the vessel is doubled and the equilibrium is allowed to be re-established, the concentration of `Q` is found to be `3M`. Find: (`a`)`K_(c)` (`b`) Concentration of `R` at two equilibrium stages. |
| Answer» (`a`) `K_(c)=(1)/(12)litre^(2)mol^(-2)`, (`b`) `4M`, `1.5 M` , | |
| 882. |
Which statement about the given reaction is incorrect? `N_(2)+3H_(2)hArr2NH_(3)`, `DeltaH=-ve`A. At `200^(@)C`, the yield of `NH_(3)` is `15%`B. At `500^(@)C`, the yield of `NH_(3)` is `15%`C. At `1000^(@)C`, the yield of `NH_(3)` is `1%`D. The reaction occurs at faster rate to attain equilibrium earlier at `500^(@)C` |
| Answer» No doubt yield is poor at `500^(@)C` but equilibrium is attained earlier. At `200^(@)C`, yield of `NH_(3)` is `88%` | |
| 883. |
Following two equilibrium is simultaneously established in a container. `PCI_5(g)hArrPCI_3(g)+CI_2(g)` `CO(g)+CI_2(g)hArrCOCI_2(g)` If some Ni(s) is introduced in the container forming `Ni(CO)_4(g)` then at new equilibrium.A. `PCI_3` concentration will increase.B. `PCI_3` concentration will decrease.C. `CI_2 concentration will remain same.D. `PCI_3` concentration will increase. |
|
Answer» Correct Answer - B |
|
| 884. |
`COCl_(2)` gas dissociates according to the equation, `COCl_(2)hArrCO(g)+Cl_(2)(g)`. When heated to 700 K the density of the gas mixture at 1.16 atm and at equilibrium is `1.16 g//litre` The degree of dissociation of `COCl_(2)` at 700K is :A. `0.28`B. `0.50`C. `0.72`D. `0.42` |
|
Answer» Correct Answer - c |
|
| 885. |
The dissociation of phosgene, which occurs according to the reaction `COCI_(2)(g)hArrCO(g)+CI_(2)(g)` Is an endothermic process. Which of the following will increase the degree of dissociation of `COCI_(2)?`A. adding `CI_(2)` to the systemB. Adding helium to the system at constant pressureC. Decreasing the temperature of the systemD. Reducing the total pressure |
|
Answer» Correct Answer - B::D (A) Backward shifting will take place. (B) Forward shifting will take place. (C) Backward shifting will take place. (D) Forward shifting will take place. |
|
| 886. |
`2CaSO_4(s)hArr2CaO(s)+2SO_2(g)+O_2(g),DeltaH gt 0` Above equilibrium is established by taking some amount of `CaSO_4(s)` in a closed container at 1600K. Then which of the following may be correct option?A. Moles of CaSO(s) will increase with increase in temperature.B. If the volume of the container is doubled at equilibrium then partical pressure of `SO_2(g)` will change at new equilibriumC. If the volume of the container is halved pressure of `O_2(g)` at new equilibrium will remain same.D. If two moles of the He gas is added at constant pressure then the moles of CaO(s) will increase. |
|
Answer» Correct Answer - A::C::D |
|
| 887. |
`2CaSO_(4)(s)hArr22CaO(s)+2SO_(2)(g)+O_(2)(g), DeltaHgt0` Above equilibrium is established by taking some amout of `CaSO_(4)(s)` in a closed container at `1600K` Then which of the following may be correct option.A. Moles of `CaO(s)` will increase with the increase in temperatureB. If the voulme of the container is doubled at equilibrium then partial pressure of `SO_(2)(g)` will change at new equilibriumC. If the volume of the container is halved partial pressure of `O_(2)(g)` at new equilibrium will remain sameD. If two moles of the He gas is added at constant pressure then the moles of `CaO(s)` will increase. |
|
Answer» Correct Answer - A::C::D (A) As reaction is endothermic therefore it will go in the forward direction hence moles of `CaO` will increase. (B) With the increase or decrease of volume partial pressure of the gases will remain same. (C) Due to the addition of inert gas at constant pressure reaction will proceed in the direction in which more number of gaseous moles are formed. |
|
| 888. |
`0.0755 g` of selenium vapours occupying a volume of `114.2 mL` at `700^(@)C` and `185 mm` of `Hg`. The vapours are in equilibrium as: `Se_(6(g))hArr3Se_(2(g))` Calculate: (`i`) Degree of dissociation of `Se`, (`ii`) `K_(p)`, (`iii`) `K_(c)`. Atomic weight of `Se` is `79`. |
| Answer» (`i`) `59%`, (`ii`) `0.168 atm`, (`iii`) `2.633xx10^(-5) mol^(2) litre^(-2)` , | |
| 889. |
An amount of solid `NH_4HS` is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure.Ammonium hydrogen sulphide decomposes to yield `NH_3` and `H_2S` gases in the flask.When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for `NH_4HS` decomposition at this temperature is :A. 0.11B. 0.17C. 0.18D. `0.30` |
|
Answer» Correct Answer - A |
|
| 890. |
The exothermic formation of `CIF_(3)` is represented by the equation: `CI_(2)(g)+2F_(2)(g)hArr2CIF_(3)(g),Delta_(r)H=-329J` which of the following will increase the quantity of `CIF_(3)` in an equilibrium mixture of `CI_(2),F_(2) "and" CIF_(3)`.A. Adding `F_(2)`B. Increasing the volume of containerC. Removing `CI_(2)`D. Increasing the temperature |
|
Answer» Correct Answer - A `CI_(2)(g)+3F_(2)(g)hArr2CIF_(3)(g),DeltaH=-329KJ`. Favourable conditions: (i) Decrease in temperature. (ii) Addition of reactants, (iii) Increase in pressure i.e., decrease in volume. |
|
| 891. |
An amount of solid `NH_4HS` is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure.Ammonium hydrogen sulphide decomposes to yield `NH_3` and `H_2S` gases in the flask.When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for `NH_4HS` decomposition at this temperature is :A. `0.11`B. `0.17`C. `0.18`D. `0.30` |
|
Answer» Correct Answer - A `NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g)` `{:("Initial presens",0,0.5,0),("At equi.",0,0.5+x,x):}` Total pressure`=0.5+2x=0.84 therefore x=0.17` atm `K_(P)=P_(NH_(3))xxP_(H_(2)S=0.11 atm^(2)`. |
|
| 892. |
If the `E^(@)` for a given reaction has a negative value, then which of the following gives the correct relationship for the of `DeltaG^(@)` and `k_(aq)`?A. `DeltaG^(ɵ) gt 0, K_(eq) lt 1`B. `DeltaG^(ɵ) gt 0, K_(eq) gt 1`C. `DeltaG^(ɵ) lt 0, K_(eq) gt 1`D. `DeltaG^(ɵ) lt 0, K_(eq) lt 1` |
|
Answer» Correct Answer - A `DeltaG=-nFE^(ɵ)` or `E^(@)=(-DeltaG)/(nF)` `:. DeltaG^(ɵ) gt 0` and `DeltaG^(ɵ)=-nRT In K_(p)` `:. K_(eq) lt1` |
|
| 893. |
For the reaction `PCl_(3)(g)+Cl_(2)(g)rarrPCl_(5)(g),K_(c) "is" 26 "at" 250^(@). K_(p)` at the same temperature is `(R=8.314JK^(-1)mol^(-1))`A. `4.6xx10^(3)`B. `5.7xx10^(3)`C. `6.0xx10^(-3)`D. `8.3xx10^(-3)` |
| Answer» Correct Answer - C | |
| 894. |
One mole of pure ethyl alcohil was treated with one mole of pure acetic acid at `25^(@)C` One -third of the acid changes into ester at equilibrium . The equilibrium constant for the reaction will be:A. `(1)/(4)`B. 2C. 3D. 4 |
|
Answer» Correct Answer - a |
|
| 895. |
One mole of ethanol is treated with one mole of ethanoic acid at `25^(@)C`. Half of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will beA. `1`B. `2`C. `3`D. `4` |
|
Answer» Correct Answer - A `{:(,C_(2)H_(5)OH,+,CH_(3)COOH,hArr,CH_(3)COOC_(2)H_(5),+,H_(2)O),("Initial",1,,1,,0,,0),("At equ.",1-1/2,,1-1/2,,1/2,,1/2):}` `K=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([C_(2)H_(5)OH][CH_(3)COOH])=(1/2xx1/2)/(1/2xx1/2)=1` |
|
| 896. |
Given below are the values of `DeltaH^(ɵ)` and `DeltaS^(ɵ)` for the reaction given below at `27^(@)C`. `SO_(2)(g)+1/2O_(2)(g) rarr SO_(3)(g)` `DeltaH^(ɵ)=-98.32 kJ mol^(-1), DeltaS^(ɵ)=-95 kJ mol^(-1)` Find `K_(p)` for the reaction |
|
Answer» Correct Answer - A::B::D `log_(10) K_(p)=(DeltaH^(ɵ))/(2.303RT)+(DeltaS^(ɵ))/(2.303xxR)` `:. log_(10)K_(p)=98320/(2.303xx8.314xx300)-95/(2.303xx8.314)` `:. K_(p)=14xx10^(12) "atm"^(-1//2)` |
|
| 897. |
A 1.0 L evacualated tank is charged with HI(g) to a pressure of 1.0 atm at 793 K. Some of the HI(g) forms `H_2(g)` and `I_2(g)` according to the equilibrium : `2HI(g)hArr H_2(g)+I_2(g) " " K_p=0.016` What is the pressure (in atm) of HI at equilibrium ?A. 0.11B. 0.13C. `0.80`D. 1.6 |
|
Answer» Correct Answer - C |
|
| 898. |
What is the minimum mass of `CaCO_3(s)`, below which it decomposes completely, required to establish equilibrium in a 6.50 litre container for the reaction : `CaCO_3(s)hArr CaO(s)+CO_2(g), K_c=0.05`A. `32.5g`B. `24.6g`C. `40.9g`D. `8.0g` |
|
Answer» Correct Answer - A `K_(c)=[CO_(2)]=0.05` mole//litre so moles of `CO_(2)=6.50xx0.05 "moles"=0.3250` moles `CaCO_(2)hArrCaO+CO_(2)` `1 "mole of" CO_(2)=1 "mole of" CaCO_(3)` `0.3250 "moles of" CO_(2)=0.3250 "moles of" CaCO_(3)=0.3250xx100 g "of" CaCO_(3)=32.50 "gm of" CaCO_(3)` |
|
| 899. |
What is the minimum mass of `CaCO_3(s)`, below which it decomposes completely, required to establish equilibrium in a 6.50 litre container for the reaction : `CaCO_3(g)hArr CaO(g)+CO_2(g), K_c=0.05`A. 32.5 gB. 24.5 gC. 40.9 gD. 8.0 g |
|
Answer» Correct Answer - A |
|
| 900. |
In the preparation of CaO from `CaCO_(3)` using the equilibrium, `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` `K_(p)` is expressed as `log K_(p)=7.282-8500/T` For complete decomposition of `CaCO_(3)`, the temperature in celsius to be used is:A. `1167`B. `894`C. `8500`D. `850` |
|
Answer» Correct Answer - B On complete decomposition of `CaCO_(3)` `log K_(p)=0` therefore `8500/T=7.282` or `T=8500/7.282=894^(@)C` |
|