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3 years ago

According to Graham’s law, the rate of effusion of a gas is inversely proportional to

Chemistry

1 answer:

VLD [36.1K]3 years ago

5 0

C. the square root of the mass of the particles.

<h3>Further explanation </h3>

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or

the effusion rates of two gases = the square root of the inverse of their molar masses:

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

$\rm M_1\times r_1^2=M_2\times r_2^2$

From this equation shows that the greater the mass of the gas, the smaller the effusion rate of the gas and vice versa, the smaller the mass of the gas, the greater the effusion velocity.

So if both gases are at the same temperature and pressure, the above formula can apply

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. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

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<u>Answer:</u> The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

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The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

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Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

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