急）在一个密闭容器中,盛有N2和H2,它们的起始浓度分别是1.8MOL/·l-1和5.4mol在一个密闭容器中,盛有N2和H2,它们的起始浓度分别是1.8MOL/·l-1和5.4mol·L,在一定的条件下它们反应生成NH3,10min后测得N

急）在一个密闭容器中,盛有N2和H2,它们的起始浓度分别是1.8MOL/·l-1和5.4mol在一个密闭容器中,盛有N2和H2,它们的起始浓度分别是1.8MOL/·l-1和5.4mol·L,在一定的条件下它们反应生成NH3,10min后测得N
急）在一个密闭容器中,盛有N2和H2,它们的起始浓度分别是1.8MOL/·l-1和5.4mol
在一个密闭容器中,盛有N2和H2,它们的起始浓度分别是1.8MOL/·l-1和5.4mol·L,在一定的条件下它们反应生成NH3,10min后测得N的浓度是0.8mol·L-1,则在这10min内NH3的平均反应速率是 ()
A.0.1mol·L-1·min-1
B.0.3mol·L-1·min-1
C.0.2mol·L-1·min-1
D.0.6mol·L-1·min-1
希望有人帮忙,理由 越详细越好.
怎么计算法?
以后遇到这种题该怎么判断
急,一天内一定采纳.

急）在一个密闭容器中,盛有N2和H2,它们的起始浓度分别是1.8MOL/·l-1和5.4mol在一个密闭容器中,盛有N2和H2,它们的起始浓度分别是1.8MOL/·l-1和5.4mol·L,在一定的条件下它们反应生成NH3,10min后测得N
怎么出来化学了
这题好算
写出方程式N2+3H2=可逆=2NH3
带入算就行了
N2 + 3H2=可逆= 2NH3
1 2
1.8-0.8 x
1/2=(1.8-0.8)/x ,得到10min后氨的浓度为2mol/L
这就算出平均反应速率2/10=0.2mol·L-1·min-1
于是答案选C

我觉得是0.2
N2的浓度1.8-0.8=1
每个NH3含一个N,一个N2含2个N原子,所以NH3的浓度是2,总共10分钟
2/10=0.2

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C

方程 N2 + 3H2 = 2NH3
起始 1.8 5.4 0
结束 0.8
反应 1 3 2（反应浓度是与系数成比例的）
所以N2的速率是V=1/10=0.1mol/(l·min)
因为速率和系数成正比
所以NH3的速率是0.2mol·L-1·min-1
选C
一般做这类题目都是用这种方法，列上面的化学方...

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方程 N2 + 3H2 = 2NH3
起始 1.8 5.4 0
结束 0.8
反应 1 3 2（反应浓度是与系数成比例的）
所以N2的速率是V=1/10=0.1mol/(l·min)
因为速率和系数成正比
所以NH3的速率是0.2mol·L-1·min-1
选C
一般做这类题目都是用这种方法，列上面的化学方程式能理清思路
望采纳。谢谢

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写出方程式N2+3H2=可逆=2NH3
带入算就行了
N2 + 3H2=可逆= 2NH3
1 2
1.8-0.8 x
1/2=(1.8-0.8)/x ，得到10min后氨的浓度为2mol/L
这就算出平均反应速率2/10=0.2mol·L-1·min-1
于是答案选C