A => B
ΔG = -12.9-Kj – (-11.3-Kj) = -1.6 Kj
ΔG = -RT∙lnKeq => lnKeq = -(ΔG/RT) =
-(-1.6Kj∙molˉ¹/0.008314Kj∙molˉ¹∙Kˉ¹∙293K) = 0.6568
Keq = exp(0.6568) = 1.928
Qeq = [2.41M]/[1.35M] = 0.5602
Keq = 1.928 > Qeq = 0.5602 => Reaction shifts right.
…………….A.............<==>.............B
…………2.41M……………....…..1.35M
……………-x……………………… +x
………2.41 – x…………………1.35 + x
Keq = [B]/[A] = [1.35+x]/[2.41+x] = 1.928
=> 1.35+x = 1.928(2.41+x) = 4.648 – 1.928x
=> 2.928x = 3.298 => x = 1.126M
Ceq: [A] = (2.41-1.126)M = 1.284M
………[B] = (1.35+1.126)M = 2.476M
Verified => Keq = [B]/[A] = (2.476)/(1.284) = 1.928 QED
Consider the system:
A (aq) → B (aq)
at 293 K where ΔGo of formation A = -11.3 kJ/mol and ΔGo standard of formation B = -12.9 kJ/mol. Calculate the
concentration of A at equilibrium when 2.41 mol of A and 1.35 mol of B are mixed in water to form exactly one liter of solution.
1 answer
1 answer