Given that ΔGo for a reaction is 4.00 kJ mol-1, what is the value of Keq...

Given that ΔGo for a reaction is 4.00 kJ mol-1, what is the value of Keq at 25oC (i.e., 77oF)? The value of R is 0.008314 kJ/mol K or 0.001986 kcal/mol K. Report your answer to three significant digits.

Solutions

Expert Solution

ΔGo= standard Gibb's free energy

R = gas constant

T = absolute temperature

K = equilibrium constant

∆G° is called the standard Gibb’s free energy, where the naught specifies a standard set of reaction conditions that include constant pressure, a given temperature, and a set of standard-state concentrations.

ΔGo= -RTlogKeq

= Keq=antilog(-ΔGo/RT)

1 J/mol = 0.000239 kcal/mol

Where ΔGo= 4 kcal/mol

= 4/0.000239

= 16736J/mol

= 16.736kJ/mol
and T= 25 oC = 298 K

Keq= antilog (-(16.736 KJ/mol)/ (0.008314 kJ/mol *298)

Keq= antilog (-16.736/ 2.477572)

Keq= antilog (-6.7550004601278994111977371394252)

Keq = 1.742* 10^ (-7)

queen_honey_blossom answered 1 year ago

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