Question

EXPLAIN PLENSE 4. Consider the reaction which is at equilibrium: Fe203(s) +2 H2(g) List all ways to drive the equilibrium to the right. 2 Fe(s) +3 H20(g), AH > 0 5. A chemist makes a saturated solution of Ga S and determines [Ga3Tley to be 0.00542 M. What is (4 pts.) the value,for Ksp of Ga2S3 at this 6. A certain weak acid has a Ka of 3.82 x 104 What is the pH and % ionization of a 6.00 M solution?

Answer 1

4)

Fe2O3(s) + 2 H2(g) --------------------- 2 Fe(s) + 3 H2O(g)           
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H 0

a) when the concentrationn of reactants are increases , the equilibrium shifts to the right side becuase the increased concentrations of reactants are decreased towords tothe right side.

b) The pressure is decreases on equlibrium conditon, the equilibrium shifts towords to the number of moles increased side, i,e right side. because in the forword direction number of moles are increases.

[ 2 ---------- 3, only gases substances can be considered]

c) The forword reaction is endothermic reaction because
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H 0. Endothermic reactions are favoured at high temperatures. SO high temperature is applied on reaction, the equlibrium shifts towords to the right side,

5)

[Ga+3] = 0.00542 M

Ga2S3(s) ---------------- 2 Ga+3(aq) + 3 S-2(aq)

                                     2S                  3S                           where S= solubility

Ksp = [Ga+3]^2 [ S-3]^3

Ksp = (2S)^2 ( 3S)^3

Ksp = 108 S^5

Ksp = 108 x ( 0.00542)^5

Ksp = 5.05x10^-10

6)

Ka= 3.82x10^-4

C= 6.00M

for weak acids

[H+] = square root of Ka xC

[H+] = square root of ( 3.82x10^-4X6.00)

[H+] = 4.79x10^-2M

-log[H+]= -log( 4.79x10^-2)

PH= 1.32

[H+] = C $\alpha$

$\alpha$ = [H+]/C = 4.79x10^-2 /6.00 =0.798x10^-2

$\alpha$ % = 0.00798%