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denis-greek [22]

3 years ago

8

Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) A reaction mixture is made containing an initial [SO2Cl2] of 2.2×10−2M

. At equilibrium, [Cl2]= 1.3×10−2M .

1 answer:

lukranit [14]3 years ago

5 0

I think you want to ask about Keq. At equilibrium, we can know [SO2Cl2] is 2.2*10-2 M -1.3*10-2M=9*10^-3 M. And [SO2]=[Cl2]. So the Keq=1.88*10^-2.

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<u>Answer:</u> The vapor pressure of the solution is 43.55 mmHg

<u>Explanation:</u>

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$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For water:</u>

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Putting values in equation 1, we get:

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<u>For water:</u>

$\chi_{\text{water}}=\frac{n_{\text{water}}}{n_{\text{water}}+n_{\text{ethyl alcohol}}}$

$\chi_{water}=\frac{1.944}{4.118}=0.472$

<u>For ethyl alcohol:</u>

$\chi_{\text{ethyl alcohol}}=\frac{n_{\text{ethyl alcohol}}}{n_{\text{water}}+n_{\text{ethyl alcohol}}}$

$\chi_{\text{ethyl alcohol}}=\frac{2.174}{4.118}=0.528$

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To calculate the vapor pressure of the solution, we use the law given by Dalton, which is:

$P_T=\sum_{i=1}^n (p_i\times \chi_i)$

Or,

$P_T=[(p_{\text{water}}\times \chi_{\text{water}})+(p_{\text{ethyl alcohol}}\times \chi_{\text{ethyl alcohol}}$

We are given:

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Vapor pressure of ethyl alcohol = 61.2 mmHg

Putting values in above equation, we get:

$p_T=[(23.8\times 0.472)+(61.2\times 0.528)]\\\\p_T=43.55mmHg$

Hence, the vapor pressure of the solution is 43.55 mmHg

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