Question

Consider the equilibrium A(g) ⇀↽ 2 B(g) + 3 C(g) at 25◦C. When A is loaded into a cylinder at 9.13 atm and the system is allowed to come to equilibrium, the final pressure is found to be 16.89 atm. What is ∆G ◦ r for this reaction?

Answer 1

Answer:

-14.60 kJ/mol

Explanation:

Equation for the reaction

A -------------->   2B   +   3C

The ICE Table is shown as:

                     A -------------->   2B     +      3C  

Initial             9.13                   0                0

Change         - x                     + 2x          + 3x

Equilibrium    9.13-x                2x              3x

(9.13-x)+2x+3x = 16.89

9..13 - x +5x  = 16.89

9.13+4x = 16.89

4x = 16.89-9.13

4x = 7.76

x = $\frac{7.76}{4}$

x = 1.94

Equilibrium pressures are as follows:

A = 9.13 -x

= 9.13 - 1.94

= 7.19 atm

B = 2x

= 2 (1.94)

= 3.88 atm

C = 3x

= 3(1.94)

= 5.82 atm

$K_p=\frac{[P_3]^2[P_c]^3}{[P_a]}$

$K_p=\frac{[3.88]^2[5.82]^3}{[7.19]}$

$K_p=412.77$

$\delta G^0_{rxn} = -RTInK_p$

$\delta G^0_{rxn} = -8.314*10^{-3}*298 In(412.77)$

$\delta G^0_{rxn} = -14.60 kJ/mol$

Answer 2

The ∆G ◦ r for this reaction is 14.60 kJ/mol.

What is pressure?

The force applied perpendicular to an object's surface per unit area across which that force is spread is known as pressure.

Given,

The reaction

A ------>   2B   +   3C

The ICE Table is

                    A ----->   2B     +      3C  

Initial             9.13                   0                0

Change         - x                     + 2x          + 3x

Equilibrium    9.13-x                2x              3x

(9.13-x) + 2x + 3x = 16.89

9.13 - x + 5x  = 16.89

9.13 + 4x = 16.89

4x = 16.89-9.13

4x = 7.76

x =$\dfrac{7.76}{4}$

x = 1.94

Now, Equilibrium pressures are

A = 9.13 -x

A = 9.13 - 1.94

A = 7.19 atm.

B = 2x

B = 2 (1.94)

B = 3.88 atm

C = 3x

C = 3(1.94)

C = 5.82 atm

$Kp =\dfrac{[P_3]^2[P_c]^3}{[P_a]} \\\\Kp= \dfrac{[3.88]^2[5.82]^3}{[7.19]}= 412.77\\\\\\\sigma G^\circ_r_x_n = -RTInKp\\\sigma G^\circ_r_x_n =-8.314\times 10^-3 \times 298In(412.77)\\\sigma G^\circ_r_x_n =-14.60\;KJ/mol$

Thus, The ∆G ◦ r for this reaction is 14.60 kJ/mol.

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