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3 years ago

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Two compounds with general formulas A2X and AX3 have Ksp=1.5×10−5M. Part A Which of the compounds has the higher molar solubilit

y? Which of the compounds has the higher molar solubility? A2X AX3

2 answers:

DiKsa [7]3 years ago

8 0

Hey there!:

To get to the molar solubility from the Ksp, you have to think about the how Ksp is calculated, and what it means.

For a compound that forms 2 ions, Ksp = X^2 where X is the molar solubility. For a compound that forms 3 ions, Ksp = 4X^3, where again, X is the molar solubility.

If you calculate the molar solubilities of each of your compounds, you will see that A2X has the higher molar solubilty

Hope this helps!

vovikov84 [41]3 years ago

5 0

Answer:

A₃X

Explanation:

In order to find the molar solubility (S) of a compound, we will use an ICE Chart.

A₂X

Let's consider the solution of A₂X.

A₂X(s) ⇄ 2 A⁺(aq) + X²⁻(aq)

I 0 0

C +2S +S

E 2S S

The solubility product (Kps) is:

Kps = 1.5 × 10⁻⁵ = [A⁺]².[X²⁻] = (2S)².S = 4S³

S = 0.016 M

A₃X

Let's consider the solution of A₃X.

A₃X ⇄ 3 A⁺(aq) + X³⁻(aq)

I 0 0

C +3S +S

E 3S S

The solubility product (Kps) is:

Kps = 1.5 × 10⁻⁵ = [A⁺]³.[X³⁻] = (3S)³.S = 27 S⁴

S = 0.027 M

A₃X has a higher molar solubility than A₂X.

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bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

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We know that,

The relation between the $C_p\text{ and }C_v$ for an ideal gas are :

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As we are given :

$C_p=28.253J/K.mole$

$28.253J/K.mole-C_v=8.314J/K.mole$

$C_v=19.939J/K.mole$

Now we have to calculate the entropy change of the gas.

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

$\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J$

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. $(w=-pdV)$

(C) Heat during the process will be,

$q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J$

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

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Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

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