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3 years ago

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2 SO2(g) + O2(g) 2 SO3(g) Assume that Kc = 0.0680 for the gas phase reaction above. Calculate the corresponding value of Kp for

this reaction at 84.5°C.

1 answer:

son4ous [18]3 years ago

6 0

Answer: The corresponding value of $K_p$ for this reaction at 84.5°C is 0.00232

Explanation:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

Relation of with is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

= equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = 0.0680

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $84.5^0C=(273+84.5)K=357.5K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=0.0680\times (0.0821\times 357.5)^{-1}\\\\K_p=0.00232$

Thus the corresponding value of $K_p$ for this reaction at 84.5°C is 0.00232

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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

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<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

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For the given balanced chemical equation:

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We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

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

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$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

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