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3 years ago

Estimate the molar mass of a gas that effuses at 1.80 times the effusion rate of carbon dioxide. answer in units of g/mol.

1 answer:

3 0

From Grahams Law the rates of effusion of two gases are inversely proportional to the square roots of their molar masses at the same temperature and pressure.
Therefore; R1/R2 = √mm2/√mm1
The molecular mass of Carbon dioxide is 44 g
Hence; 1.8 = √(44/x
3.24 = 44/x
x = 44/3.24
= 13.58
Therefore, the molar mass of the other gas is 13.58 g/mol

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Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of $NO$ = $-\frac{1d[NO]}{2dt}$ = 0.066 M/s

Rate in terms of disappearance of $O_2$ = $-\frac{1d[O_2]}{dt}$

Rate in terms of appearance of $NO_2$ = $\frac{1d[NO_2]}{2dt}$

1. The rate of formation of $NO_2$

$-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}$

$\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s$

2. The rate of disappearance of $O_2$

$-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}$

$-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s$

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