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2 years ago

2 points

Chemistry

1 answer:

quester [9]2 years ago

6 0

(b) 12.7 L

<u>Explanation:</u>

Given:

Volume, V₁ = 12 L

Pressure, P₁ = 760 mm Hg (at STP)

Temperature, T₁ = 273 K (at STP)

Pressure, P₂ = 720 mm Hg

Temperature, T₂ = 0°C = 273 K

Volume, V₂ = ?

We know:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

Substituting the value we get:

$\frac{760 X 12}{273} = \frac{720 X V_2}{273} \\\\V_2 = \frac{760 X 12}{720} \\\\V_2 = 12.7 L$

Therefore, V₂ is 12.7 L

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Answer :

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The equilibrium constant of this reaction is, 0.0216

The equilibrium pressure of O₂(g) is, 0.147 atm

Explanation :

The given chemical reaction is:

$PCl_3(l)\rightarrow PCl_3(g)$

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

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{M(s)}\times \Delta G^0_{(M(s))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]-[n_{M_3O_4(s)}\times \Delta G^0_{(M_3O_4(s))}]$

where,

$\Delta G^o$ = Gibbs energy of reaction = ?

n = number of moles

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

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (0kJ/mol)]-[1mole\times (-9.50kJ/K.mol)]$

$\Delta G^o=9.50kJ/mol$

The Gibbs energy of reaction is, 9.50 kJ/mol

Now we have to calculate the equilibrium constant of this reaction.

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = 9.50kJ/mol = 9500 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K = equilibrium constant = ?

$9500J/mol=-(8.314J/K.mol)\times (298K)\times \ln (K)$

$K=0.0216$

The equilibrium constant of this reaction is, 0.0216

Now we have to calculate the equilibrium pressure of O₂(g).

The expression of equilibrium constant is:

$K=(P_{O_2})^2$

$0.0216=(P_{O_2})^2$

$P_{O_2}=0.147atm$

The equilibrium pressure of O₂(g) is, 0.147 atm

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