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The reaction between gaseous NH_(3) and HBr produces the white solid NH_(4)Br. Suppose that NH_(3) and HBr are introduced simultaneously into the opposite ends of an open tube that is 1 metre long. Where would you expect the white solid to form ? |
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Answer» Solution :Suppose the two gases meet to form a white solid `NH_(4)Br` at a will of `r_(1)`cm from the `NH_(3)` END. Thus from the HBR end, the distance will be `(100 - r_(1)) cm`. We have, `(r_(NH_(3)))/(r_(HBr)) = SQRT((M_(HBr))/(M_(NH_(3)))) = sqrt((81)/(17))` Since the rate of diffusion is proportional to the distance, the molecules travel. `therefore (r_(NH_(3)))/(r_(HBr)) = (r_(1))/(100 - r_(1)) = sqrt((81)/(17)) = 2.18` `therefore r_(1) = 65.55 cm`. |
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