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101.

In a reaction `A+B hArr C+D` , the initial concentrations of A and B were 0.9 mol `dm^(-3)` each. At equilibrim the concentration of D was found to be 0.6 mol `dm^(-3)`. What is the value of equilibruim constant for the reaction ?A. 8B. 4C. 9D. 3

Answer» Correct Answer - B
`{:(,A,+,B,hArr, C,+,D),("Initial conc.",0.9M,,0.9M,,-,,-),("At equilibrium",0.3M,,0.3M,,0.6M,,0.6M):}`
`K=([C][D])/([A][B])=(0.6xx0.6)/(0.3xx0.3)=4`
Discussion
102.

The equilibrium constant `(K)` of a reaction may be written asA. `K=e^(-DeltaG//RT)`B. `K=e^(-DeltaG^(@)RT)`C. `K=e^(-DeltaH//RT)`D. `K=e^(-DeltaH^(@)RT)`

Answer» Correct Answer - B
`Delta G^(@)=-nE^(@)F ` and
`E^(@)=(RT)/(nF) log_(e) K`
`:. DeltaG^(@)=-RT log_(e)K`
`:. K=e^(-DeltaG^(@)//RT)`
Discussion
103.

1 mol of `N_(2)` and 2 mol of `H_(2)` are allowed to react in a 1 `dm^(3)` vessel. At equilibrium, `0.8` mol of `NH_(3)` is formed. The concentration of `H_(2)` in the vessel isA. 0.6 moleB. 0.8 moleC. 0.2 moleD. 0.4 mole

Answer» Correct Answer - B
`{:(,N_(2),+,3H_(2),hArr , 2NH_(3)),("Initial ",1 "mol",,2"mol",,),("At eq.",1-0.4=0.6"mol",,2-3xx0.4=0.8"mol",,0.8"mol"):}`
Formation of 0.8 mol of `NH_(3)` means of 0.4 mol `N_(2)` have reacted
Discussion
104.

5 moles of `SO_(2)`and 5 moles of `O_(2)` are allowed to react .At equilibrium , it was foumnd that `60%` of `SO_(2)` is used up .If the pressure of the equilibrium mixture is one aatmosphere, the parital pressure of `O_(2)` is :A. 0.82 atmB. 0.52 atmC. 0.21 atmD. 0.41 atm

Answer» Correct Answer - D
`{:(,2SO_(2)(g),+,O_(2)(g),hArr,2SO_(3)(g)),("Initially",5,,5,,0),("Used up",(5xx60)/(100)=3,,3xx(1)/(2)=1.5,,1.5xx2=3),("At eqm. ",5-3=2,,5-1.5=3.5,,):}`
`:. p_(O_(2))=("Moles of" O_(2))/("Total no. of moles" )xx"Total pressure "`
`=(3.5)/(2+3.5+3)xx1=(3.5)/(8.5)xx1=0.41 "atm."`
Discussion
105.

Assertion (A) `:` For gaseous reaction when `Delta n =0 , K_(p)=K_(c)`. `Deltan =` change in the number of gas moles Reason (R ) `:` For gaseous reaction, `K_(p)=K_(c)(RT)^(Deltan)`A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true but R is not a correct explanation of A.C. A is true but R is falseD. A is false but R is true.

Answer» Correct Answer - a
Discussion
106.

Solubility of NaOH in waterA. increases with increase in temperatureB. decreases with increase in temperatureC. is not affected by a change in temperatureD. first increases and then decreases with temperature.

Answer» Correct Answer - B
Dissolution of NaOH in water is an exothermic reaction
`NaOH +` Water `hArr NaOH` solution `+` Heat
As such increasing temperature shifts the equilibrium in the backward direction thus decreasing the solubility
Discussion
107.

Solubility of `N_(2)` in waterA. increases with increase in temperatureB. decreases with increase in temperatureC. is not affected by a change in temperatureD. first increases and then decreases with temperature.

Answer» Correct Answer - B
Solubility of all gases in water is exothermic `( Delta H ` is `-ve )` . As such solubility of `N_(2)` in water decreases with increase in temperature.
Discussion
108.

For the hypothetical reaction `A +B hArr C+2D` equilibrium constant at 400 K is `1.8 xx 10^(-6)` mol `L^(-1)`A. `9.5 xx 10^(-5) mol L^(-1)`B. `9.5 xx 10^(-4) mol L^(-1)`C. `4.75 xx10^(-4) mol L^(-1)`D. None of these

Answer» Correct Answer - B
`{:(,A,+,B,hArr,C,+,2D),("Initial conc." (mol L^(-1)),-,,-,,1.0,,1.0),("At equilibrium" (mol L^(-1)),((1-0-x)/(2)),,((1.0-x)/(2)),,((1.0+x)/(2)),,x):}`
D used up `=(1.0 -x )` mol `L^(-1)`
C used up `=((1.0-x)/(2)) mol L^(-1)`
C at equilibrium `(1.0-(1.0-x)/(2))mol L^(-1)`
`=((1.0+x)/(2)) mol L^(-1)`
A ( or B ) produced at equilibrium `=C` used up
`=((1.0-x)/(2))mol L^(-1)`
A very small value of K indicates that at equilibrium concentrations of C and D are very small i.e., x is very small. As such at equilibrium
`[C]=((1.0+x)/(2))=0.5 mol L^(-1)`
`[D]=x mol L^(-1) , [A]=[B]`
`=((1.0-x)/(2)) mol L^(-1)`
`=0.50 mol L^(-1)`
`:. K=([C][D]^(2))/([A][B])`
`1.8 xx 10^(-6)=(0.50xx x^(2))/(0.50 xx 0.80)`
`x^(2)=9.0 xx 10^(-9)`
`implies x=9.5 xx10^(-4) mol L^(-1)`
`:. [D]=x=9.5 xx 10^(-4) mol L^(-1)`
Discussion