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Pepsi [2]
3 years ago
8

Consider the following mechanism.

Chemistry
1 answer:
solong [7]3 years ago
7 0

Answer :

(a) The overall equation is:

ClO^-(aq)+I^-(aq)\rightarrow Cl^-(aq)+IO^-(aq)

(b) The intermediates are  :

OH^-(aq),HClO(aq)\text{ and }HIO(aq)

Explanation :

<u>Part (a) :</u>

(1) ClO^-(aq)+H_2O(l)\rightarrow HClO(aq)+OH^-(aq)   (fast)

(2) I^-(aq)+HClO(aq)\rightarrow HIO(aq)+Cl^-(aq)    (slow)

(3) OH^-(aq)+HIO(aq)\rightarrow H_2O(l)+IO^-(aq)   (fast)

By adding the three equations and cancelling the common terms on both side, we will get the overall equation.

ClO^-(aq)+I^-(aq)\rightarrow Cl^-(aq)+IO^-(aq)

<u>Part (b)  :</u>

Intermediates are generated and consumed in the mechanism and do not include in the overall equation.

Since, intermediates will not include in the overall mechanism.

The intermediates are  :

OH^-(aq),HClO(aq)\text{ and }HIO(aq)

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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
padilas [110]

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

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

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)]

For the given equation:

\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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Volume of the reaction vessel is increased - shift to the left

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Explanation:

When a constraint such as a change in temperature, pressure or volume is imposed on a reaction system in equilibrium, the equilibrium position will shift in such a way as to annul the constraint.

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Since the reaction is exothermic (heat is given out) when the reaction is cooled down, the forward reaction is favoured.

Adding of reactants shifts the equilibrium position to the right hand side hence when H2 is added, the equilibrium position shifts to the right.

Decreasing the pressure shifts the equilibrium position to the direction of higher total volume hence the equilibrium shifts to the left when pressure is decreased.

A catalyst has no effect on the equilibrium position. It increases the rate of forward and reverse reaction to the same extent hence the equilibrium position is unaffected.

Removal of water from the system increases the rate of forward reaction since a product is being removed from the reaction system.

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