For a chemical reaction `2X + Y rarrZ`, the rate of appearance of `Z` is `0.05 mol L^(-1) min^(-1)`. The rate of diappearance of `X` will be

Answer

0.1

Explanation

Rate of reaction
`=-1/2 (d[X])/(dt) = -(d[Y])/(dt) = + (d[Z])/(dt)`……. (i)
`therefore` Rate of appearance of
`z = (d[Z])/(dt)= 0.05 mol L^(-1)` per min.
`therefore` Rate of disappearance of `x = -(d[X])/(dt)`
From equation (i), we get
`-1/2(d[X])/(dt) = (d[Z])/(dt), therefore -(d[X])/(dt) = 2 xx (d[Z])/(dt)`
`=2 xx 0.05 = 0.1 mol L^(-1) min^(-1)`