Question

In a particular experiment, equal moles of H2(g) at 1.00 atm and Br2(g) at 1.00 atm were mixed in a 1.00 L flask at 25°C and allowed to reach equilibrium. Then the molecules of H2 at equilibrium were counted using a very sensitive technique, and 1.10 ✕ 1013 molecules were found. For this reaction, calculate the values of ΔS°, ΔG°, and K.

Answer 1

Answer:

∆S = -348.3Jmol^-1 K^-1

K = 4

∆G ~= - 3435.3JK^-1mol^-1

Explanation:

H2 (g) + Br2(g) ----->2HBr(g)

x x 2x. Before eq

Given that ∆H(reaction) =103.8KJ/mol

and that At equilibrium the number of hydrogen molecules found is;

1.10 × 10¹³

Therefore we can say

Number(H2 molecules = Number (Bromine molecules)

And according to the reaction ratio, Number(HBr molecules) = 2 × 1.10 × 10¹³

T= 25°C = 273 + 25 = 298K

∆H = T∆S............(1)

∆S = ∆H/T

∆S =( -103.8KJ/mol/)/298K

∆S = - 0.3483KJmol^-1K^-1

∆S = -348.3Jmol^-1 K^-1

K = [HBr]²/[H2][Br2]

K = [2×1.10 × 10¹³]²/[1.10× 10¹³][1.10×10¹³]

K = (2.20)² ×10^26/[1.21 × 10^26]

K = 4.84/1.21

K = 4

∆G = -RTlnK = -2.303RTlog10(K)

Where R is the universal gas constant; R = 8.314JK^-1mol^-¹

∆G = - 2.303 ×8.314 × 298 × log(4)

∆G ~= - 3435.3JK^-1mol^-1

Answer 2

The values for:

∆S° = -348.3 $Jmol^{-1} K^{-1}$

K = 4

∆G° = - 3435.3 $JK^{-1}mol^{-1}$

Balanced chemical reaction:

                                            H₂ (g) + Br₂(g) ----->2HBr(g)

Before equilibrium:                x              x                2x

Given:

∆H(reaction) =103.8KJ/mol

$K_{eq}$ = 1.10 × 10¹³

Number(H₂ molecules) = Number (Bromine molecules)

And according to the reaction ratio,

Number(HBr molecules) = 2 * 1.10 * 10¹³

T= 25°C = 273 + 25 = 298K

Solving for Entropy change:

∆H = T∆S............(1)

∆S = ∆H/T

∆S =( -103.8KJ/mol/)/298 K

∆S = - 0.3483 K$Jmol^{-1} K^{-1}$

∆S = -348.3 $Jmol^{-1} K^{-1}$

K = [HBr]²/[H₂ ][Br₂]

K = [2×1.10 × 10¹³]²/[1.10× 10¹³][1.10×10¹³]

K = (2.20)² ×10^26/[1.21 × 10^26]

K = 4.84/1.21

K = 4

∆G° = -RTlnK

∆G°= -2.303RTlog10(K)

where R is the universal gas constant; R = 8.314 $JK^{-1}mol^{-1}$

∆G° = - 2.303  * 8.314 * 298 * log(4)

∆G° = - 3435.3$JK^{-1}mol^{-1}$

Find more information about Entropy change here:

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