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標題:

equilibrium

發問:

Consider the following equilibrium.2 NOBr(g) 2 NO(g) + Br2(g) If nitrosyl bromide, NOBr, is 34 percent dissociated at 25°C and the total pressure is 0.25 atm, calculate Kp and Kc for the dissociation at this temperature.

最佳解答:

• Probability_5
• 一個男人 35, 對結婚的看法-@1@
• 夾時晨八字結婚 (於2015年4月30結婚)

• Probability_5
• 一個男人 35, 對結婚的看法-@1@
• 夾時晨八字結婚 (於2015年4月30結婚)

 

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2NOBr(g) = 2NO(g) + Br2(g) Initially: No. of moles of NOBr = y mol No. of moles of NO = No. of moles of Br2 = 0 mol At eqm: No. of moles of NOBr = y - (34%)y = 0.66y mol No. of moles of NO = (34%)y = 0.34y mol No. of moles of Br2 = (34%)y/2 = 0.17y mol Total number of moles of all components = 0.66y + 0.34y + 0.17y = 1.17y PNOBr = XNOBrPT = (0.66/1.17) x 0.25 = 0.25 x (66/117) atm PNO = XNOPT = (0.34/1.17) x 0.25 = 0.25 x (34/117) atm PBr2 = XBr2PT = (0.17/1.17) x 0.25 = 0.25 x (17/117) atm Kp = (PNO)2 (PBr2) / (PNOBr)2 = [0.25 x (34/117)]2 [0.25 x (17/117)] / [0.25 x (66/117)]2 = 0.00964 atm (or 974 Pa, as 1 atm = 1.01 x 105 Pa) Concentration = n/V = P/RT For NOBr: PNOBr = 0.25 x (66/117) atm = 0.25 x (66/117) x (1.01x105) Pa R = 8.314 J mol-1K-1 T = 273 + 25 = 298 K [NOBr]eq = PNOBr/RT = = [0.25 x (66/117) x (1.01x105)] / (8.314 x 298) = 5.75 mol m-3 Similarly, [NO]eq = PNO/RT = [0.25 x (34/117) x (1.01x105)] / (8.314 x 298) = 2.96 mol m-3 [Br2]eq = PBr2/RT = [0.25 x (17/117) x (1.01x105)] / (8.314 x 298) = 1.48 mol m-3 Hence. Kc = [NO]eq2 x [Br2]eq / [NOBr]eq2 = (2.96)2 (1.48) / (5.75)2 = 0.392 mol m-3 = 0.000392 mol dm-3

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