Consider the following reaction resulted from gas grill barbeque over a Labor\'s

Question

Consider the following reaction resulted from gas grill barbeque over a Labor's day holiday: 1- Predict the sign of delta S degree before calculating its value/explain delta S degree is negative 2- Calculate the delta S degree and delta G degree from the given data at standard conditions of temperature and pressure? Given the following reaction: NOCl_2(g) rightarrow NO_(g) + Cl_2(g) This reaction has delta H degree _rxn = 50.0 Kj/mol and delta S degree _rxn = 100.0 J/(mol. K). a- Is the reaction exothermic or endothermic reaction? b- Is the reaction spontaneous at all temperatures? If not explain c- Calculate the temperature at which the reaction becomes spontaneous?

Explanation / Answer

9) So = Soproduct - Soreactant

So = [ (2 x 214.0 J/mol.K) + ( 13 x 70 J/mol.K) - [ ( 8 x 150 J/mol. K ) + ( 10 x 125 J/mol K) ]

So = 1313 J/mol.K - 2450 J/mol.k

So = -1137 J/mol.K

so, So is negetive

now we find Ho

Ho = Hoproduct- Horeactant

Ho = [ ( 2 x -394.0 KJ/mol) + (13 x -286.0 KJ/mol) ] - [ (8x -104 KJ/mol ) + ( 10 x 0 ) ]

Ho = -4506 KJ/mol + 832 KJ/mol

Ho = -3674 KJ/mol

Go = Ho - T So .......{ T = temperature in kelvin standard temperature = 273.15 K }

Go = -3674000 J/mol - (273.15 k x -1137 J/mol.K )

Go = -3363428.45 J/mol = -3363.43 KJ/mol

10) Ho is positive so reaction is endothermic  

Go = Ho - T So

Go = 50000 J/mol - (T x 100 J/mol.K)

for spontaneous reaction Go will be negetive ( Go < 0 ) so,

in this case both So and Ho is positive

Go < 0 only at high temperature   

T > 500 K

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