N2O4 is 25% dissociated at 37°C and 1 atm, calculate the percentage dissociation at 0.1 atm and 37° C. (A) 63.2% (B) 36.6% (C) 65.1% (D) 46.6% Share with your friends Share 17 Vartika Jain answered this Dear Student, For this we need to calculate Kp for the reactionN2O4 ↔ 2NO2 1 0 1-x 2xTotal moles=1-x+2x=1+xpN2O4 = 1-x1+x×PIt is given that x=0.25 and P=1 atmSo, pN2O4=1-0.251+0.25×1=0.6 atmpNO2=2x1+x×PpNO2=2×0.251+0.25×1=0.4 atmSo, Kp =(pNO2)2(pN2O4)=(0.4)2(0.6)=0.267 atmNow, let the degree of dissociation at 0.1 atm be ypN2O4 = 1-y1+y×P=1-y1+y×0.1pNO2=2y1+y×P=2y1+y×0.1Kp = (pNO2)2(pN2O4)=(2y1+y×0.1)2(1-y1+y×0.1)=0.267 atmor, 0.667y2=0.276 atmor, y=0.632 % y=63.2%Hence, correct answer is (A) 47 View Full Answer
