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Would you expect the following reactions to be spontanious at high temperatures,

ID: 840100 • Letter: W

Question

Would you expect the following reactions to be spontanious at high temperatures, low temperatures, or not at all? Explain your choice.

1)PCl3(g) + Cl2(g) --> PCl5(s) H* = -87.9 kJ

2) 2 N2O(g) ---> 2N2(g) + O2(g) H* = -164.1 kJ

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The molar mass of aluminum sulfite Al2(SO3)3 is 294.15 g/mol.

1) How man moles of Al2(SO3)3 are in 4.23 grams of Al2(SO3)3?

2) How many alumminum ions are present in .637 moles of Al2(SO3)3? (please show work)

3) How many grams of oxygen are present in a 1.25 mole sample of Aluminum sulfite? (please show work)

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=How many mL of a .200 M methanol solution are needed to make 150 mL of a 0.0375 M methanol solution? (please show work so I can understand how it was done)

Explanation / Answer

1)PCl3(g) + Cl2(g) --> PCl5(s) H* = -87.9 kJ

A reaction is spontaneous if dG < 0

we know that

dG = dH - TdS

Given dH < 0

now look at the reaction

calculate dn

dn = moles of gaseous products - moles of gaseous reactants

dn = 1-2

dn = -1

As dn <0 , dS < 0

for the reaction to be spontaneous


dH - TdS < 0

As dH < 0 and dS <0

is possible only at low values of T

so the reaction is spontaneous at low temperatures


2) 2 N2O(g) ---> 2N2(g) + O2(g) H* = -164.1 kJ

A reaction is spontaneous if dG < 0

we know that

dG = dH - TdS

Given dH < 0

now look at the reaction

calculate dn

dn = moles of gaseous products - moles of gaseous reactants

dn = 3-2

dn = 1

As dn > 0 , dS > 0

for the reaction to be spontaneous


dH - TdS < 0

As dH < 0 and dS > 0

is possible only at any values of T

so the reaction is spontaneous at all temperatures

3) moles = mass /molar mass

moles = 4.23 / 294.15

moles = 0.01438


Al2(S03)3 ---> 2 Al3+ + 3 S032-

so

moles of Al+3 = 2 x moles of Al2(S03)3

moles of Al+3 = 2 x 0.637

moles of Al+3 = 1.274

Given Al2(S03)3

mass of oxygen in 1 mole of Al2(S03)3 = 9 x 16 = 144

mass of oxygen in 1.25 mole of Al2(S03)3 = 144 x 1.25 = 180 g

4) apply M1V1 = M2V2


V1 x 0.2 = 0.0375 x 150

V1 = 28.125

so 28.125 ml is required

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