For the reaction SO3(g) + NO(g) NO2(g) + SO2(g) Kc was found to be 0.500 at...

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For the reaction SO3(g) + NO(g) NO2(g) + SO2(g) Kc was found tobe 0.500 at a certain temperature. A reaction mixture is preparedin which 0.130 mol NO2 and 0.130 mol of SO2 are placed in a 3.00 Lvessel. After the system reaches equilibrium what will be theequilibrium concentrations of all four gases?

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4.4 Ratings (752 Votes)

the reaction is SO3(g) + NO ----> NO2(g)+SO2(g)

Kc = [SO2] [NO2]/ [SO3]*[NO]=0.5

[SO3][NO2]/ ]SO2] [NO2] =1/0.5= 2

since the coefficients in the numerator and denominator are same. wther one takes moles or concentrations does not make any difference

MolesÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â SO3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â NOÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â NO2Â Â Â Â Â Â Â Â Â Â Â SO2

IniritialÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.13Â Â Â Â Â Â Â Â Â Â Â Â 0.13

At eq.Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.13-xÂ Â Â Â Â Â Â Â Â 0.13-x

x= moles of SO3 formed at equilibrium

Kc= 2= [SO3] [NO]/ [SO2][NO2] = x²/(0.13-x)² =2

taking square root gives

x/(0.13-x)= sqrt(2)= 1.44

x= 1.44*(0.133-x)

2.44x= 1.44*0.133

x= 0.0785

At equilibfrium [SO3] =[NO] =0.0785/3 =0.026 [SO2] =(0.13-0.0.0785)/3=0.017167

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