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951.

The dissociation of ammonium carbamate may be represents bt the equation, `NH_4CO_2NH_2(s)hArr2NH_3(g)+CO_2(g)` `DeltaH^circ`for the forward reaction is negative, The equilibrium will shift from right to left if there is :A. a decrease in pressureB. an increase in temperatrureC. an increase in the concentration of ammoniaD. an increase in the concentration of carbon dioxde

Answer» Correct Answer - B::C::D
Discussion
952.

For a reversible reaction `aA+bBhArrcC+dD`, the variation of `K` with temperature is given by log `(K_(2))/(K_(1))=(-DeltaH^(@))/(2.303R) [(1)/(T_(2))-(1)/(T_(1))]` then,A. `K_(2)gtK_(1) "if" T_(2)gtT_(1)` for an endothermic changeB. `K_(2)ltK_(1) "if" T_(2)gtT_(1)` for an endothermic changeC. `K_(2)gtK_(1) "if" T_(2)gtT_(1)` for an endothermic changeD. `K_(2)gtK_(1) "if" T_(2)gtT_(1)` for an endothermic change

Answer» Correct Answer - A::D
Discussion
953.

For a reaction, `aA+bBhArrcC+dD`, the reaction quotient `Q=([C]_(0)^(c)[D]_(0)^(d))/([A]_(0)^(a)[B]_(0)^(b))`, where `[A]_(0)`, `[B]_(0)`, `[C]_(0)`, `[D]_(0)` are initial concentrations. Also `K_(c)=([C]^(c)[D]^(d))/([A]^(a)[B]^(b))` where `[A]`, `[B]`, `[C]`, `[D]` are equilibrium concentrations. The reaction proceeds in forward direction if `Q lt K_(c)` and in backward direction if `Q gt K_(c)`. The variation of `K_(c)` with temperature is given by: `2303log(K_(C_(2)))/(K_(C_(1)))=(DeltaH)/(R)[(T_(2)-T_(1))/(T_(1)T_(2))]`. For gaseous phase reactions `K_(p)=K_(c)(RT)^(Deltan)` where `Deltan=` moles of gaseous products `-` moles of gaseous reactants. Also `-DeltaG^(@)=2.303RT log_(10)K_(c)`. Which relation is correct?A. `2.303 log_(10)K= -(DeltaH^(@))/(RT)+(DeltaS^(@))/(R )`B. `DeltaG=DeltaG^(@)+2.303RT log_(10)Q`C. `K=Ae^(-DeltaH^(@)//RT)`D. all are correct

Answer» All are correct.
`DeltaG^(@)=-2.303Rtlog_(10)K_(c)` (At eqm. `DeltaG=0`)
`DeltaH^(@)-TDeltaS^(@)=-2.303RT log_(10)K`
`2.303 log_(10)K_(c)=(DeltaH^(@))/(RT)-(DeltaS^(@))/(R )`
Discussion
954.

What is the effect of temperature and pressure on the yields of products? a. `N_(2)(s)+3H_(2)(g) hArr 2NH_(3)+x cal` b. `N_(2)(g)+O_(2)(g) hArr 2NO(g)-y cal` c. `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)+46.9 kcal` d. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)-15.0 kcal`A. I, IIB. II, IIIC. I, IIID. III, IV

Answer» Correct Answer - 3
Discussion
955.

`aA+bBhArrcC+dD` In above reaction low pressure and high temperature, conditions are shift equilibrium in back direction so correct set:A. `(a+b)gt(c+d),DeltaHgt0`B. `(a+b)lt(c+d),DeltaHgt0`C. `(a+b)lt(c+d),DeltaHlt0`D. `(a+B)gt(c+d),DeltaHlt0`

Answer» Correct Answer - D
`aA+bBhArrcC+dD`
At high temp. & low pressure equilibrium is shifting in backward direction. It means `(a+b)gt(c+d)` & heat will reaction in the formation of producers is `DeltaHlt0`.
Discussion
956.

In the formation of nitric acid, `N_(2)` and `O_(2)` are made to combine. Thus, `N_(2)+O_(2) hArr 2NO+"Heat"` which of the following condition will favour the formation of NO?A. low temperatureB. high temperatureC. freezing pointD. all are favourable

Answer» Correct Answer - A
`N_(2)+O_(2) hArr 2NO+"Heat"`
The above equation shows that the reaction is endothermic. Endothermic reactions are forward in forward direction at high temperature.
Discussion
957.

If the temperature of the system at equilibrium is increased, the equilibrium will shift in the direction which ………… heat.

Answer» Correct Answer - Absorbs
Discussion
958.

For the reaction, `H_(2)(g)+CO_(2)(g) hArr CO(g)+H_(2)O(g)`, if the initial concentration of `[H_(2)]=[CO_(2)]` and x mol `L^(1)` of `H_(2)` is consumed at equilibrium, the correct expression of `K_(p)` is:A. `x^(2)/((1-x)^(2))`B. `((1+x)^(2))/((1-x)^(2))`C. `(1+x^(2))/((2+x)^(2))`D. `x^(2)/(1+x^(2))`

Answer» Correct Answer - A::B::C
`{:(,H_(2)(g),+,CO_(2)(g),hArr,CO(g),+,H_(2)O(g)),(t=0,x,,x,,0,,0),(t=t_("Eq"),1-x,,1-x,,x,,x):}`
`:. K=([CO][H_(2)O])/([H_(2)][CO_(2)])=x^(2)/((1-x)^(2))`
Discussion
959.

For the reaction, `H_(2)(g)+CO_(2)(g) hArr CO(g)+H_(2)O(g)`, if the initial concentration of `[H_(2)]=[CO_(2)]` and x mol `L^(-1)` of `H_(2)` is consumed at equilibrium, the correct expression of `K_(p)` is:A. `(x^(2))/((1-x)^(2))`B. `((1+x)^(2))/((1-x)^(2))`C. `(x^(2))/((2+x)^(2))`D. `(x^(2))/(1-x^(2))`

Answer» Correct Answer - A
`{:(,H_(2(g)),+,CO_(2(g)),hArr,CO_((g)),+,H_(2)O_((g))),("Initial conc.",1,,1,,0,,0),("At equilibrium",(1-x),,(1-x),,x,,x):}`
`K_(p) = (PCO.PH_(2)O)/(PH_(2).PCO_(2)) = (x.x)/((1-x)(1-x)) = (x^(2))/((1-x)^(2))`
Discussion
960.

What concentration of `CO_(2)` be in equilibrium with `2.5xx10^(-2) mol L^(-1)` of CO at `100^(@)C` for the reaction: `FeO(s)+CO(g) hArr Fe(s)+CO_(2)(g), K_(c)=5.0`

Answer» `K_(c)=([CO_(2)])/([CO])`
`:. 5=([CO_(2)])/(2.5xx10^(-2))`
`:. [CO_(2)]` at equilibrium `=2.5xx10^(-2)xx5`
`=12.5xx10^(-2)mol litre^(-1)`
Discussion
961.

What concentration of `CO_(2)` be in equilibrium with `0.025 M` CO at `120^(@)C` for the reaction `FeO(s)+CO(g) hArr Fe(s)+CO_(2)(g)` if the value of `K_(c)=5.0` ?A. `0.125 M`B. `0.0125 M`C. `1.25 M`D. `12.5 M`

Answer» Correct Answer - A
`{:(,FeO(s),+,CO(g),hArr,Fe(s),+,CO_(2)(g)),("At eq.",-,,0.025,,-,,x),(,"solid",,,,"solid",,):}`
`K_(c)=[(CO_(2)(g))/(CO(g))]=x/0.025=5.0` (Given)
`:. x=0.125 M`
`[CO_(2)]=0.125 M`
Discussion
962.

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. `0.52 atm`B. `0.21tm`C. `0.41`aatmD. `0.82 atm`

Answer» Correct Answer - c
Discussion
963.

`CO(g)+Cl_2(g) hArr COCl(g)+Cl(g) K_(eq)=1.5xx10^(-39)` If the rate constant , K, for the forward reaction is `1.4xx10^(-28) L mol^(-1) sec^(-1)` what is K ( in L `mol^(-1) sec^(-1)` ) for the backward reaction ?A. `2.1xx10^(-67)`B. `1.0xx10^(-11)`C. `9.3xx10^(10)`D. `7.1xx10^(27)`

Answer» Correct Answer - C
Discussion
964.

`1` mol of A in `1` litre vessel maintained at constant T shows the equilibrium `A(g) hArr B(g) +2C(g) " " K_(C_(1))` `C(g) hArr 2D(g)+3B(g) " " K_(C_(2))` If the equilibrium pressure is `13/6` times of initial pressure and `[C]_(eq)=4/9[A]_(eq)`, Calculate `K_(C_(1))` and `K_(C_(2))`.

Answer» `{:(A(g),hArr,B(g)+,2C(g)and ,C,hArr,2D+,3B(g)),(1,,0,0,,,,),(1-a,,(a+3b),2a-b,2a-b,,2b,(a+3b)):}`
`:. Sigma` mole at equilibrium `=1-a+a+3b+2a-b+2b`
`=2a+4b+1`
Now, `P_("initial") prop 1`
`P_(eq) prop 2a+4b+1`
`:. 2a+4b+1=13/6`
or `2a+4b=7/6`
Also `[C]_(eq)=4/9 [A]_(eq)`
`2a-b=4/9xx(1-a)`
`22a-9b=4`
By equations (i) and (ii) `a=0.25, b=0.167`
`:. K_(C_(1))=([B][C]^(2))/([A])=((a+3b)(2a-b)^(2))/((1-a))`
`=([0.25+(3xx0.167)][0.5-0.167]^(2))/((1-0.25))=0.11`
`:. K_(C_(2))=([D]^(2)[B]^(2))/([C])=((2b)^(2)(a+3b)^(3))/((2a-b))`
`=([2xx0.167]^(2)[0.25+(3xx0167)]^(3))/([0.5-0.167])=0.142`
Discussion
965.

The equilibrium constant K for the reaction `2HI(g) hArr H_(2)(g)+I_(2)(g)` at room temperature is `2.85` and that at `698 K` is `1.4 xx10^(-2)`. This impliesA. ExothermicB. EndothermicC. ExergonicD. Unpredictable

Answer» Correct Answer - A
With the increase of temperature, k value decreases, so the forward reaction decreases with increase of temperature. This implies that reaction will proceed in forward direction with decrease of temperature, i.e., heat is librated and hence forward reaction is exthoermic.
Discussion
966.

Given the following reaction at equilibrium `N_(2)(g) + 3H_(2)(g)hArr2NH_(3)(g)`. Some inert gas at constant pressure is added to the system. Predict which of the following facts:A. more `NH_(3)`is producedB. Less `NH_(3) (g)` is producedC. No affect on the equilibriumD. `K_(p)` of the reaction is decreased

Answer» Correct Answer - b
Discussion