Q. \( \delta g = \delta g + r t \ln q \).

Q: What is the correct form of the equation \\Delta G = \\Delta G + RT\ln ?

The corrected formulis \\Delta G = \\Delta G^\circ + RT\ln . Here \\Delta G is the Gibbs free energy at nonstandard conditions, \\Delta G^\circ is the standard free energy change, R is the gas constant, T is temperature, and is the reaction quotient.

Q: How does \\Delta G relate to spontaneity?

If \\Delta G < 0 the reaction is spontaneous as written. If \\Delta G > 0 it is nonspontaneous. At equilibrium \\Delta G = 0 , so the reaction has no net driving force.

Q: How do \\Delta G^\circ and the equilibrium constant K relate?

At equilibrium = K and \\Delta G = 0 , so \\Delta G^\circ = -RT\ln K . Thus K = e^{-\\Delta G^\circ / RT} .

Q: What units should I use for R and T ?

Use consistent SI units: R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}} and T in kelvin. Ensure \\Delta G and \\Delta G^\circ are in joules per mole when using this R .

Q: What if < K or > K ?

If < K , then \\Delta G < 0 and the reaction proceeds forward to reach equilibrium. If > K , then \\Delta G > 0 and the reaction proceeds in reverse.

Q: How does temperature affect \\Delta G vithis equation?

Temperature appears explicitly in the RT\ln term and implicitly in \\Delta G^\circ . Changing T alters both terms, so reactions can switch spontaneity with temperature changes.

Answer

Correct relation: \( \Delta G = \Delta G^\circ + RT \ln Q \). Solving for the standard free energy gives \( \Delta G^\circ = \Delta G – RT \ln Q \). This follows by simple rearrangement.

Detailed Explanation

We are given the relation \( \Delta G = \Delta G^\circ + RT \ln Q \). I will solve this equation step by step and explain each algebraic operation in detail. I will first isolate the term containing \( Q \) and then solve explicitly for \( Q \).

Step 1. Subtract \( \Delta G^\circ \) from both sides to isolate the logarithmic term. This is a standard algebraic operation that keeps the equation balanced because we perform the same operation on both sides of the equality. After subtracting \( \Delta G^\circ \) we obtain:

\[
\Delta G – \Delta G^\circ = RT \ln Q
\]

Step 2. Divide both sides of the equation by \( RT \) to obtain \( \ln Q \) alone. Division by \( RT \) is valid provided \( R \neq 0 \) and \( T \neq 0 \). The gas constant \( R \) is a known nonzero constant, and \( T \) must be in kelvin and nonzero for the thermodynamic expression to be meaningful. After dividing we have:

\[
\frac{\Delta G – \Delta G^\circ}{RT} = \ln Q
\]

Step 3. Exponentiate both sides to remove the natural logarithm. The natural exponential function \( \exp(\,\cdot\,) \) is the inverse of \( \ln(\,\cdot\,) \). Applying \( \exp \) to both sides yields:

\[
\exp\!\left( \frac{\Delta G – \Delta G^\circ}{RT} \right) = \exp(\ln Q)
\]

Step 4. Use the identity \( \exp(\ln x) = x \) for \( x>0 \). The reaction quotient \( Q \) must be positive for the logarithm to be defined, so we obtain the solution for \( Q \):

\[
Q = \exp\!\left( \frac{\Delta G – \Delta G^\circ}{RT} \right)
\]

This is the explicit expression for the reaction quotient \( Q \) in terms of the Gibbs free energy \( \Delta G \), the standard Gibbs free energy \( \Delta G^\circ \), the gas constant \( R \), and the absolute temperature \( T \).

Optional. If instead you wanted to solve the original equation for \( \Delta G^\circ \), you would rearrange by subtracting \( RT \ln Q \) from both sides. That gives:

\[
\Delta G^\circ = \Delta G – RT \ln Q
\]

Both rearrangements are algebraically equivalent and are useful in different contexts. The main solution requested, solved step by step, is:

\[
Q = \exp\!\left( \frac{\Delta G – \Delta G^\circ}{RT} \right)
\]

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Chemistry FAQs

What is the correct form of the equation \( \\Delta G = \\Delta G + RT\ln \) ?

The corrected formulis \( \\Delta G = \\Delta G^\circ + RT\ln \). Here \( \\Delta G \) is the Gibbs free energy at nonstandard conditions, \( \\Delta G^\circ \) is the standard free energy change, \( R \) is the gas constant, \( T \) is temperature, and \( \) is the reaction quotient.

How does \( \\Delta G \) relate to spontaneity?

If \( \\Delta G < 0 \) the reaction is spontaneous as written. If \( \\Delta G > 0 \) it is nonspontaneous. At equilibrium \( \\Delta G = 0 \), so the reaction has no net driving force.

How do I compute \( \) for reaction?

\( \) is the ratio of product activities to reactant activities, raised to stoichiometric powers. For ideal gases or dilute solutions, use partial pressures or concentrations: \( = \frac{[C]^c[D]^d}{[A]^a[B]^b} \). Use activities for nonideal systems.

How do \( \\Delta G^\circ \) and the equilibrium constant \( K \) relate?

At equilibrium \( = K \) and \( \\Delta G = 0 \), so \( \\Delta G^\circ = -RT\ln K \). Thus \( K = e^{-\\Delta G^\circ / RT} \).

What units should I use for \( R \) and \( T \) ?

Use consistent SI units: \( R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}} \) and \( T \) in kelvin. Ensure \( \\Delta G \) and \( \\Delta G^\circ \) are in joules per mole when using this \( R \).

What if \( < K \) or \( > K \) ?

If \( < K \), then \( \\Delta G < 0 \) and the reaction proceeds forward to reach equilibrium. If \( > K \), then \( \\Delta G > 0 \) and the reaction proceeds in reverse.

How does temperature affect \( \\Delta G \) vithis equation?

Temperature appears explicitly in the \( RT\ln \) term and implicitly in \( \\Delta G^\circ \). Changing \( T \) alters both terms, so reactions can switch spontaneity with temperature changes.

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