A) Compute ?rH°, ?rS°, and ?Stotal and determine the spontaneity of this reactio

Question

A) Compute ?rH°, ?rS°, and ?Stotal and determine the spontaneity of this reaction under standard state conditions at 25 °C: Under standard state conditions at 25 °C, this reaction is: Using values from the previous problem, determine the approximate temperature that the reaction below is at equilibrium under standard state conditions, assuming no phase changes occur.?

KNO3 -494.63 Kj/mol

C 0 Kj/mol

KCN -113.0 Kj/mol

O2 0 Kj/mol

A) Compute AHo, AS°, and AStotal and determine the spontaneity of this reaction under standard state conditions at 25 °C: KNO, (e) + C (c, graphite)KcN (e) + 1.5 o, ) Number Number Number 4HP- kJ/ mol AS°- J/ Kmol AStotal- J/ Kmol Under standard state conditions at 25 °C, this reaction is: O O O Spontaneous in forward direction At equilibrium Spontaneous in reverse direction Using values from the previous problem, determine the approximate temperature that the reaction below is at equilibrium under standard state conditions, assuming no phase changes occur. KNO, e) C (e, graphite)KCN (e) + 1.5 o, ) Number equil

Explanation / Answer

Enthalpy data , delrtaH0(Kjmole): KMO3= -494.63 Kj/mole, O2=0, KCN= -113.0 Kj/mole, C=0

entropy daya, deltaS0 (J/mole.K): KNO3= 132.92, O2=205.03, KCN= 127.77, C=5.69

for the reaction KNO3(s)+C( graphite) ---------->KCN(s)+1.5O2(g)

standard enthalpy change, deltaH0= sum of standard enthalpy change of products- sum of enthalpy change of reactants

=1* standard enthalpy of KCN+1.5* standard enthalpy change of O2- (1* standard enthalpy change of KNO3+1* standard enthalpy change of C)

=1*-113+1.5*0 -(1*(-494.63)+0 = 381.63 KJ

similarly, deltaS0 = 1*127.77+1.5*205.03-(1*132.92+5.69)=296.7 J/K

deltaGo= standard gibbs free energy change= deltaHo- T*deltaS0

at 298K, deltaGo = 381.63*1000- 298*296.7 J=293213.4 J =293.213 Kj since deltaG0 is >0, the reaction is non- spontaneous.

the reaction is spontaneous when deltaG<0, Calculate the temperature at which first deltaG=0

361.63*1000-T* 296.7=0

T = 1319.9

so any temperature <1319.9, the reaction is spontaneous since at T>1319.8, the reaction becomes spontaneous.

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