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densk [106]
2 years ago
14

Q1. The equation for the reaction between sulfur dioxide and oxygen is shown.

Chemistry
1 answer:
Delicious77 [7]2 years ago
3 0

The mole fraction of sulfur trioxide in the equilibrium mixture is 0.353.
Hence option (B) is the correct answer.

<h3>What is mole fraction ?</h3>

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

We have to use algebra here,

Because in the equation O_2 has a stoichometry of 1, the other have 2,

let O_2 = x

You started with 2 moles, so the moles of O_2 at equilibrium will be 2 - X

SO_3 has a stoichiometry of 2 compared to O_2, so will be 2X

( not minus from anything as you started with no moles of SO_3)

And then SO_2  will also be 2X as it’s stoichiometry is 2, but as you started with 2 moles it will be 2-2X

If you add these in an equation to get total moles it’s

2 - 2X  +  2 - X + 2X = 3.4 (total given)

and then gathering together becomes 4-X = 3.4

so,

4 = 3.4 + X

so,

x = 0.6

Therefore,

the moles of SO_3  = 2 x 0.6 = 1.2

mole fraction = moles of the gas / total moles of gas

mole fraction = 1.2 / 3.4 = 0.353

Learn more about mole fraction here ;

brainly.com/question/25994904

#SPJ1

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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:

K_{sp}\text{ of }Ag_2S=8\times 10^{-51}

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G^o=-RT\ln K

where,

\Delta G^o = standard Gibbs free energy = ?

R = Gas constant = 8.314J/K mol

T = temperature = 25^oC=[273+25]K=298K

K = equilibrium constant or solubility product = 8\times 10^{-51}

Putting values in above equation, we get:

\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ

For the given chemical equation:

Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)

The equation used to calculate Gibbs free change is of a reaction is:  

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]

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))})]

We are given:

\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ

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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