Calculate the value of the equilibrium constant, Keq , for the following reaction at 298 Kelvin.?

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Calculate the value of the equilibrium constant, Keq , for the following reaction at 298 Kelvin.?

Calculate the value of the equilibrium constant, Keq , for the following reaction at 298 Kelvin.
(Use the reaction free energy given below.)

2NH3(g) + CO2(g) = NH2CONH2(aq) + H2O(l)

ΔGo = −55.6 kJ/mol

2007-03-28 02:18:32 · 2 answers · asked by mahbubur r 1 in Science & Mathematics ➔ Chemistry

2 answers

ΔGo = -RTlnKeq =>

Keq= e^-(ΔGo/RT) =e^-(-55600/(8.314*298)) =5.57*10^9

2007-03-28 02:42:18 · answer #1 · answered by bellerophon 6 · 1⤊ 0⤋

Yep 5.57e9 is correct.

2007-03-29 03:30:41 · answer #2 · answered by Raki 3 · 0⤊ 0⤋