1.

Consider the reaction, N_(2)(g)+3H_(2)(g) to 2NH_(3)(g) The equility relationship between (d[NH_(3)])/(dt) and -(d[H_(2)])/(dt) is

Answer»

`+(d[NH_(3)])/(dt)=-(2)/(3)(d[H_(2)])/(dt)`
`+(d[NH_(3)])/(dt)=-(3)/(2)(d[H_(2)])/(dt)`
`(d[NH_(3)])/(dt)=(d[H_(2)])/(dt)`
`+(d[NH_(3)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)`

SOLUTION :If we write rate of REACTION in terms of concentration of `NH_(3) and H_(2)`, then
Rate of reaction`=(1)/(2)(d[NH_(3)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)`
So, `(d[NH_(3)])/(dt)=-(2)/(3)(d[H_(2)])/(dt)`


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