The standard potential for the reaction,
`Ag^(+)(aq.)+Fe^(2+)(aq.)toFe^(3+)(aq.)+Ag(s)`
is 0.028V. What is the standard free energy change for this reaction?
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find out the standard free energy change at `60^(@)C`and at 1 atn if the `N_(2)O_(4)` is 50 % dissociated
Correct Answer - c `N_(2)O_(4)(g)Leftrightarrow2NO_(2)(g)` ` N_(2)O_(4)` is 50 % dissociated , the mole fraction of the substance is given by `x_(n2O4)= (1-05)/(1 + 0.5), x_(no2)= (2xx0.5)/(1+0.5)` `P_(n2O4)=0.5/1.5xx 1 atm , p_(no2)1/1.5...
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The correct relationship between free energy change in a reaction and the corresponding equilibrium constant K is
Correct Answer - A `DeltaG^(@) = - RT In K `
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The `K_(P)` for reaction `A + B iff C + D ` is 1.34 at `60^(@)C` and 6.64 at ` 100 ^(@)C` . Determine the standard free energy change of this reaction
Correct Answer - `- 810 J//mol ; - 5872 J//ml and 41.3 kJ //mol`. `DeltaG=- 2.303 RT log K` `DeltaG_(1) = - 810J//mol` `DeltaG_(2) = - 5872J//mol` `log .(K_(2))/(K_(1))= (DeltaH)/(2.303 R)[(T_(2)-T_(1))/(T_(1)T_(2))]`...
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For the water gas reaction, `C(s) +H_(2)O(g) hArr CO(g)+H_(2)(g)` the standard Gibbs free energy of reaction (at `1000K)` is `-8.1 kJ mol^(-1)`. Calcu
We know that `K="antilog"((-DeltaG^(@))/(2.303RT))` . . . .(i) Given that, `DeltaG^(@)=-8.1kJ//mol` `R=8.314xx10^(-3)kJ" "K^(-1)mol^(-1)` Substituting these values in eq. (i), we get `K="antilog"[(-(-8.1))/(2.303xx8.314xx10^(-3)xx1000)]` `K=2.65`
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The standard free energy change of a reaction is `DeltaG^(@)=-115` at 298K. Calculate the equilibrium constant `K_(P)` in log `K_(P).(R=8.314JK^(-1)mo
Correct Answer - A `log" "K_(P)=(-DeltaG^(@))/(2.303RT)` `=(-(-115xx1000))/(2.303xx8.313xx298)` `=20.16`
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Calculate the standard free energy change for the combustion of glucose at 298K, using the given data, `C_(6)H_(12)O_(6)+6O_(2)to6CO_(2)+6H_(2)O` `Del
`DeltaG^(@)=-2882.58kJ" "mol^(-1)`
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Calculate free energy change for the reaction, `H_(2)(g)+Cl_(2)(g)to2H-Cl(g)` By using the bond energy and entropy data. Bond energies of `H-H,Cl-Cl a
Correct Answer - 190.9kJ `DeltaG^(@)` can be calculated by using: `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `DeltaH^(@)=sum(BE)_("reactants")-sum(BE)_("products")` `=435+240-2xx430=-185kJ` `DeltaS^(@)=sumS_("products")^(@)-sumS_("reactants")^(@)` `=2xx186.68-130.59-222.95` `=19.82JK^(-1)=19.82xx10^(-3)kJ" "K^(-1)` `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `=-185-298xx19.82xx10^(-3)=-190.9kJ`
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The following reaction is performed at 298K `2NO(g) + O_(2)(g) hArr 2NO_(2)(g)` The standard free energy of formation of NO(g) is 86.6 KJ/mol at 298 K
Correct Answer - D `2DeltaG_(f(NO_(2)))^(@) -[2DeltaG_(f(NO))^(@)+DeltaG_(f(O_(2)))^(@)] = DeltaG_(r)^(@) =- RTlnK_(p)` `2DeltaG_(f(NO_(2))^(@) -[2xx 86,600+0] =- RTlnK_(p)` `DeltaG_(f(NO_(2))^(@) = 0.5[2xx86,600-R(298) ln(1.6 xx 10^(12))]`
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Calculate the free energy change at 298 K for the reaction, `Br_(2)(l)+ Cl_(2)(g) rarr 2BrCl(g)`. For the reaction `DeltaH^(@)=29.3` KJ & the entropie
Correct Answer - A `DeltaG-29.3 xx10^(3)-(2xx239.7 - 152.3 -223)xx298 =-1721.8 "Joule".`
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By giving energy to a hydrogen atom in energy state n=1. If the potential energy of the atom in n=1 state be `-13.6 eV`, calculate (a) potential enreg
Correct Answer - `-0.85 eV, 12.75 eV, 971 Å`
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