Ch 14 Hw https://session.masteringchemistry CHEM 106 McCoy $16 14HWExercise 14.9
ID: 986183 • Letter: C
Question
Ch 14 Hw https://session.masteringchemistry CHEM 106 McCoy $16 14HWExercise 14.97 assignmentProblemID 590381078offset-next Signed in as S Exercise 14.97 Part A A sample of pure NO, is healed to 39 Ct Caloulate Ke for the reaction. which temperaturepartaly dissociates acording to the equation 2 NO, (g) 2 NO(g) + Ola) Al equilbrium the density of the gas mixture is 0.520 g/L at 0.755 atm Kc- Submit MyAnswers up Pr Feedback hy one t equation as i stated in 1kp tc41(K2074 / COAqu2r4to gette desrednns er than ly correct equation as get the MacBook Air 80 5 6 7 8 9 0Explanation / Answer
let volume be 1L , P = 0.755 atm , T = 339+273 = 612 K ,
PV = nRT hence ( 0.755 x 1) = n x 0.08206 x 612 , n = total moles of gas = 0.015
Let initial moles of NO2 = a , then at equilibrum NO2 moles = a-2x , NO2 moles = 2x , O2 moles = x
total gas moles at equilibrium = a+x = 0.015
mass of mixture = density x volume = 0.52 x 1 = 0.52 g
Now mass = moles x molar mass
hence total mass 0.52 = ( a-2x) x 46 + ( 2x) ( 30 ) + ( X)(32)
0.52 = 46a-92x + 60x + 32x
0.52 = 46a , a = 0.0113 ,
a+x = 0.015 , x= 0.015-0.0113= 0.0037 ,
hence now Kc = [NO]^2 [O2] /[NO2] = ( 2 x 0.0037)^x ( 0.0037) / ( 0.0113-2(0.0037)^2
= 0.00333