What is the value of K for this aqueous reaction at 298 K? Delta G = 19.54 kj/mol
Given data,
Temperature = 298K
ΔG = 19.54 KJ/mol
= 19540 J/mol
Let us consider a reaction,
ΔG = -R * T * ln Kc
19540 = - 8.314 * 298.0 * ln(Kc)
ln Kc = - 19540 / 2477.572
ln Kc = -7.88
Kc = e-7.88
Kc = 3.78 * 10-4
What is the value of K for this aqueous reaction at 298 K? Delta G =...
Calculate the value of K, at 298 K, for each value of delta G degree. delta G degree = 7.5 kJ/mol delta G degree = -9.0 kJ/mol
What is the value of K for this aqueous reaction at 298 K? A + B F C + D AG° = 15.82 kJ/mol KE 0
What is the value of K for this aqueous reaction at 298 K? A+BC + D AG = 18.78 kJ/mol K = 0.00037
What is the value of ? for this aqueous reaction at 298 K? A+B↽−−⇀C+DΔ?°=14.14 kJ/mol A + B ↽ − − ⇀ C + D Δ G ° = 14.14 kJ / mol ?= K =
What is the value of K for this aqueous reaction at 298 K? A+B = C+D AG° = 19.57 kJ/mol K =
What is the value of K for this aqueous reaction at 298 K? A+B <-----> C+D deltaG= 20.59 kj/mol
What is the value of ? for this aqueous reaction at 298 K? A+B↽−−⇀C+DΔ?°=29.94 kJ/mol A + B ↽ − − ⇀ C + D Δ G ° = 29.94 kJ / mol
What is the value of K for this aqueous reaction at 298 K? A+B ΔGo 23.77 kJ/mol Number 2.57x 10 10
What is the value of K for this aqueous reaction at 298 K? A+B F C +D AG° = 20.93 kJ/mol K = At 25 °C, the equilibrium partial pressures for the reaction 3 A(g) + 2 B(g) = C(g) + 2D(g) were found to be Pa = 4.86 atm, PB = 5.54 atm, Pc = 4.91 atm, and Pb = 5.82 atm. What is the standard change in Gibbs free energy of this reaction at 25 °C? AGеxn =...
What is the value of K for this aqueous reaction at 298 K? A+B=C+D AG° = 29.82 kJ/mol K=