Free energy question involving Kp and deltaG

In summary, the problem involves calculating the reaction enthalpy and Gibbs reaction energy for the given reaction at standard conditions, as well as determining the value of Kp. The equations and values provided in the problem are used to find the necessary values, with the standard ΔG being used in the equation for ln(Kp). It is also noted that the standard heats of formation are needed to calculate the standard reaction enthalpy.
  • #1
hahaha158
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Homework Statement



The following reaction takes place at standard conditions

4NH3(g)+6NO(g)----->5N2(g)+6H20

A)calculate reaction enthalpy and Gibbs reaction energy

B)Calculate Kp


Homework Equations



the table gives the following values for deltaG0f[X]
NH3(g)=-16.5 Kj/mol
NO(g)=87.6 KJ/mol
H20(g)=-228.6 KJ/mol

ln(Kp)=-deltaG/(RT)


The Attempt at a Solution



I finished part a) and found gibbs reaction energy to be -1815 KJ

For part b) i am unsure of what to enter for deltaG in the equation ln(Kp)=-deltaG/(RT). Do I use the reaction energy calculated from part a), deltaG for NH3, NO, or H20? Any help is appreciated, thanks!
 
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  • #2
you would use the standard DG found by this formula:
∑nΔGoproducts - ΣmΔGoreactants
 
  • #3
-the n and m being the coefficients for each molecule
 
  • #4
hahaha158 said:

Homework Statement



The following reaction takes place at standard conditions

4NH3(g)+6NO(g)----->5N2(g)+6H20

A)calculate reaction enthalpy and Gibbs reaction energy

B)Calculate Kp

Homework Equations



the table gives the following values for deltaG0f[X]
NH3(g)=-16.5 Kj/mol
NO(g)=87.6 KJ/mol
H20(g)=-228.6 KJ/mol

ln(Kp)=-deltaG/(RT)

The Attempt at a Solution



I finished part a) and found gibbs reaction energy to be -1815 KJ

For part b) i am unsure of what to enter for deltaG in the equation ln(Kp)=-deltaG/(RT). Do I use the reaction energy calculated from part a), deltaG for NH3, NO, or H20? Any help is appreciated, thanks!
You use the ΔG0 that you calculated in part a (I didn't check your arithmetic). (Also, that should not be a ΔG in the equation; it should be ΔG0). To get the standard reaction enthalpy, you need to know the standard heats of formation. Did they give you these?

Chet
 

FAQ: Free energy question involving Kp and deltaG

What is Kp and how is it related to free energy?

Kp, or the equilibrium constant, is a measure of the ratio of products to reactants at equilibrium in a chemical reaction. It is related to free energy through the equation ΔG = -RTlnKp, where R is the gas constant and T is the temperature in Kelvin.

How is the change in free energy (ΔG) related to the spontaneity of a reaction?

The change in free energy, ΔG, is a measure of the energy available to do work in a system. If ΔG is negative, the reaction is considered spontaneous and can occur without any outside energy input. If ΔG is positive, the reaction is non-spontaneous and requires energy input to occur.

Can Kp and ΔG be used interchangeably to determine the spontaneity of a reaction?

No, Kp and ΔG are not interchangeable. Kp is a measure of the equilibrium state of a reaction, while ΔG is a measure of the overall energy change in a reaction. A reaction with a negative ΔG is always spontaneous, but a reaction with a high Kp may not necessarily be spontaneous.

How does temperature affect Kp and ΔG?

The equilibrium constant, Kp, is directly affected by temperature, with an increase in temperature causing an increase in Kp. The change in free energy, ΔG, is also affected by temperature through the equation ΔG = ΔH - TΔS, where ΔH is the change in enthalpy and ΔS is the change in entropy. As temperature increases, the contribution of ΔS increases and can lead to a more negative ΔG, making the reaction more spontaneous.

Can Kp and ΔG be used to predict the direction of a reaction?

Yes, Kp and ΔG can both be used to predict the direction of a reaction. A negative ΔG and a high Kp both indicate that the reaction will proceed in the forward direction. A positive ΔG and a low Kp indicate that the reaction will proceed in the reverse direction. However, other factors such as concentration and catalysts must also be considered when predicting the direction of a reaction.

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