The decomposition of N2O4(g) is described by the followingequilibrium. The initi

Question

The decomposition of N2O4(g) is described by the followingequilibrium. The initial pressure inside a flask which containsonly N2O4(g) is 1.00 atm; at equilibrium 20% of the N2O4(g) hasdecomposed. N2O4(g)<--------> @ NO2(g) Write the expression fro Kp and calculate the value ofKp. If the original pressure of N2O4(g) was only .010 atm,calculate the % decompostion at equilibrium. The decomposition of N2O4(g) is described by the followingequilibrium. The initial pressure inside a flask which containsonly N2O4(g) is 1.00 atm; at equilibrium 20% of the N2O4(g) hasdecomposed. N2O4(g)<--------> @ NO2(g) Write the expression fro Kp and calculate the value ofKp. If the original pressure of N2O4(g) was only .010 atm,calculate the % decompostion at equilibrium.

Explanation / Answer

Given that                                          N2O4(g)   <--------> 2 NO2(g) Initial(atm)          1.0                                0 Change(atm)      -x                                +2x Equi(atm)           1.0-x                            2x Given that at equilibrium 20% of the N2O4(g) has decomposed thatmeans 80 % of N2O4(g) remains.     1.0 -x = 0.8 atm               x = 0.2 atm            Kp = (2x)2 / (1.0-x)                  = (2*0.2)2 / (0.8)                  = 0.2                        N2O4(g)   <--------> 2 NO2(g) Initial(atm)          0.01                                0 Change(atm)      -x                                +2x Equi(atm)         0.01-x                            2x                     Kp = (2x)2 / (0.01-x)                    0.2 = (2x)2 / (0.01-x) On solving above equation x = 0.00854 atm                    the % decompostion at equilibrium = (0.00854 atm/0.01atm)*100                                                                         = 85.4 %                        N2O4(g)   <--------> 2 NO2(g) Initial(atm)          0.01                                0 Change(atm)      -x                                +2x Equi(atm)         0.01-x                            2x                     Kp = (2x)2 / (0.01-x)                    0.2 = (2x)2 / (0.01-x) On solving above equation x = 0.00854 atm                    the % decompostion at equilibrium = (0.00854 atm/0.01atm)*100                                                                         = 85.4 % Initial(atm)          0.01                                0 Change(atm)      -x                                +2x Equi(atm)         0.01-x                            2x                     Kp = (2x)2 / (0.01-x)                    0.2 = (2x)2 / (0.01-x) On solving above equation x = 0.00854 atm                    the % decompostion at equilibrium = (0.00854 atm/0.01atm)*100                                                                         = 85.4 %

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