If you find that the standard Gibbs free energy change for a reaction is 3.26 kJ

Question

If you find that the standard Gibbs free energy change for a reaction is 3.26 kJ/molK, what is the equilibrium constant for the reaction at 25.0 oC? Be careful with your units in the calculation.



For a reaction, a plot of ln K versus inverse temperature is found to be linear and fits the following equation: y = -1,693x + 7.46.  What is the value of ?H?  Give your answer with units of kJ/molK.
If you find that the standard Gibbs free energy change for a reaction is 3.26 kJ/molK, what is the equilibrium constant for the reaction at 25.0 oC? Be careful with your units in the calculation.



For a reaction, a plot of ln K versus inverse temperature is found to be linear and fits the following equation: y = -1,693x + 7.46.  What is the value of ?H?  Give your answer with units of kJ/molK.
If you find that the standard Gibbs free energy change for a reaction is 3.26 kJ/molK, what is the equilibrium constant for the reaction at 25.0 oC? Be careful with your units in the calculation.



For a reaction, a plot of ln K versus inverse temperature is found to be linear and fits the following equation: y = -1,693x + 7.46.  What is the value of ?H?  Give your answer with units of kJ/molK.

Explanation / Answer

Delta G= - 2.303 RTlog Keq

Delta G= 3.26 kJ/molK

Keq= ?

Temperature T= 25 oC = 298 K

Log K= 3.26 X1000/2.303X298X 8.314 = 0.571343614

K= 3.72

SLOPE = Delta H /2.303 R = -1693

Delta H = -1693x 2.303 x 8.314 = 32.416 kJ/molK

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