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某次考试中有一道题说: 在某高温下,NO_2分解 O_2)达到平衡时,NO摩尔数占混和气体总摩尔数的40％。 (1)求平衡时NO的摩尔数。 (2)求平衡时NO_2和O_2的摩尔数分别占混和气体总摩尔数的百分率。 这道题没有给定NO_2的起始摩尔数和反应器的容积,但指出了反应温度是一定的,则反应的平衡常数K为一定值。考生能不能任意假设NO_2的起始摩尔数,从而求出NO的平衡摩尔数呢?本文仅对这个问题加以讨论。 假设NO_2的起始摩尔数为n(n>0),平衡时NO的摩尔数为x、x符合题意,平衡时混和
One question in an examination stated: At a certain high temperature, when NO 2 decomposes and O 2 reaches equilibrium, the number of NO moles accounts for 40% of the total moles of gas mixture. (1) The number of moles of NO at equilibrium. (2) The percentage of the moles of NO 2 and O 2 in the equilibrium is the percentage of the total moles of the gas mixture. The initial number of moles of NO 2 and the volume of the reactor are not given in this question, but it is pointed out that the reaction temperature is constant and the equilibrium constant K of the reaction is a certain value. Can the examinee arbitrarily assume the initial number of moles of NO 2 and find the equilibrium molar number of NO? This article only discusses this issue. Assume that the initial number of moles of NO 2 is n (n>0), and the number of moles of NO at equilibrium is x and x corresponds to the meaning of the problem.