`DeltaH_(f)^(@)` for `HgO` is `-21.7 kcal//mol` at `25^(@)C`. Calculate the mass of mercury which can be liberated by treatment of excess of `HgO` with `10.0 kcal. At. Wt.` of `Hg=200.6`.
`DeltaH_(f)^(@)` for `HgO` is `-21.7 kcal//mol` at `25^(@)C`. Calculate the mass of mercury which can be liberated by treatment of excess of `HgO` with `10.0 kcal. At. Wt.` of `Hg=200.6`.
Text Solution
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Given `Hg+1//2Orarr HgO, DeltaH^(@)=-21.7kcal`
`:. HgOrarr Hg+1//2O,DeltaH^(@)+21.7kcal,`
`21.7kcal` heat gives `200.6g Hg`
`10.0 kcal `heat gives `(200xx10)/(21.7)=92.44g Hg`
`:. HgOrarr Hg+1//2O,DeltaH^(@)+21.7kcal,`
`21.7kcal` heat gives `200.6g Hg`
`10.0 kcal `heat gives `(200xx10)/(21.7)=92.44g Hg`
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Calculate the mass of mercury which can be liberated from HgO at 25^(@) C by the treatment of excess of HgO with 41.85 KJ of heat at (a) constant pressure and (b) constant volume conditions : Given DeltaH_(f)^(o) [HgO(s)] = –90.8 KJ/mole, M(Hg) = 200.6 gm/mole.
Calculate the mass of mercury which can be liberated from HgO at 25^(@)C by the treatment of excess HgO with 41.84 kJ of heat at : (a) constant pressure (b) constant volume Given : Delta H_(f)^(@)(HgO, s) = -90.8 kJ mol^(-1) & M(Hg) = 200.6 g mol^(-1) .
Knowledge Check
Equilibrium constant for a reaction can be obtained by kinetics approach or by thermodynamics approch. While from kinetics approach at equililbrium, the rate of forward and backward will be same, from thermodynamics approach Gibbs free energy will be minimized at equilibrium. Using this information and following thermodynamics values, answer the question that follow: DeltaG_(f)^(@)A(g) =-200kcal/mole DeltaG_(f)^(@)B(g) =-320 kcal/mole DeltaG_(f)^(@)C(g) =-300kcal/mole DeltaG_(f)^(@)D(l) =-224.606 kcal/mole DeltaG_(f)^(@)D(g) = -226.9.9 kcal/ mole, All values at 500K Calculate rate constant of the backward reaction for the following reaction at 500K: A(g)+B(g)iffC(g)+D(l) " if " K_(f)=10"bar"^(-1) sec^(-1)
Equilibrium constant for a reaction can be obtained by kinetics approach or by thermodynamics approch. While from kinetics approach at equililbrium, the rate of forward and backward will be same, from thermodynamics approach Gibbs free energy will be minimized at equilibrium. Using this information and following thermodynamics values, answer the question that follow: DeltaG_(f)^(@)A(g) =-200kcal/mole DeltaG_(f)^(@)B(g) =-320 kcal/mole DeltaG_(f)^(@)C(g) =-300kcal/mole DeltaG_(f)^(@)D(l) =-224.606 kcal/mole DeltaG_(f)^(@)D(g) = -226.9.9 kcal/ mole, All values at 500K Calculate rate constant of the backward reaction for the following reaction at 500K: A(g)+B(g)iffC(g)+D(l) " if " K_(f)=10"bar"^(-1) sec^(-1)
A
`10 "bar"^(-1) sec^(-1)`
B
`0.1 "bar"^(-1) sec^(-1)`
C
`0.1 sec^(-1)`
D
`10 sec^(-1)`
Equilibrium constant for a reaction can be obtained by kinetics approach or by thermodynamics approch. While from kinetics approach at equililbrium, the rate of forward and backward will be same, from thermodynamics approach Gibbs free energy will be minimized at equilibrium. Using this information and following thermodynamics values, answer the question that follow: DeltaG_(f)^(@)A(g) =-200kcal/mole DeltaG_(f)^(@)B(g) =-320 kcal/mole DeltaG_(f)^(@)C(g) =-300kcal/mole DeltaG_(f)^(@)D(l) =-224.606 kcal/mole DeltaG_(f)^(@)D(g) = -226.9.9 kcal/ mole, All values at 500K Calculate equilibrium concentration of B(g) if A(g) at 10 bar, B(g) at 2 bar,C(g) at 20 bar is mixed with excess liquid D such that following equilbrium gets established at 500K: A(g) B(g)iffC(g)+D(g)
Equilibrium constant for a reaction can be obtained by kinetics approach or by thermodynamics approch. While from kinetics approach at equililbrium, the rate of forward and backward will be same, from thermodynamics approach Gibbs free energy will be minimized at equilibrium. Using this information and following thermodynamics values, answer the question that follow: DeltaG_(f)^(@)A(g) =-200kcal/mole DeltaG_(f)^(@)B(g) =-320 kcal/mole DeltaG_(f)^(@)C(g) =-300kcal/mole DeltaG_(f)^(@)D(l) =-224.606 kcal/mole DeltaG_(f)^(@)D(g) = -226.9.9 kcal/ mole, All values at 500K Calculate equilibrium concentration of B(g) if A(g) at 10 bar, B(g) at 2 bar,C(g) at 20 bar is mixed with excess liquid D such that following equilbrium gets established at 500K: A(g) B(g)iffC(g)+D(g)
A
2M
B
`(2)/(41.57)`
C
`1M`
D
`(1)/(41.57)`
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