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
