One mole of N2O4 (g) at 300 K is kept in a closed container under one atmospheric pressure. It is heated to 600 K when 20% by mass of N2O4 (g)decomposes to NO2(g). The resultant pressure is???

Asked by Yash Mishra | 30th Nov, 2013, 09:48: AM

Expert Answer:

                                                                        N2O4(g)                                2NO2(g)

At start                                                 100/92 = 1.08mol                                 0

At equilibrium                                       80/92 = 0.86                                 20/46 mol= 0.43 mol

                                      According to ideal gas equation, at two condition

At 300K                           P0V = n0RT0

                                         1 × V = 1.08 × R × 300      -------------- (i)

At 600K                        P1V = n1RT1

                                      P1 × V  = (0.86 + 0.43) × R ×600 ------------(ii)

Divide (ii) by (i)

P1  =  1.29  ×  600

           1.08 × 300

P1 = 1.29  ×  2  =  2.38 atm

           1.08

 

Answered by Vaibhav Chavan | 2nd Dec, 2013, 09:18: AM

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