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Chapter 53: Problem 11

If \(\Delta H^{\circ}=T \Delta S^{\circ}\), what is the value of \(\Delta G^{\circ}\) ? Predict the value of \(K\) in this case.

Short Answer

Expert verified
The calculated value for \(\Delta G^{\circ}\) is 0 and hence, the predicted value for \(K\) is 1.

Step by step solution

01

Substitute the given into the Gibbs-Helmholtz equation

Start from the given formula: \(\Delta H^{\circ}=T \Delta S^{\circ}\). This can be substituted into the Gibbs-Helmholtz equation which is: \(\Delta G^{\circ} = \Delta H^{\circ}-T \Delta S^{\circ}\). The substitution will simplify the equation.
02

Solve for \(\Delta G^{\circ}\)

By replacing the value of \(\Delta H^{\circ}\) into the Gibbs-Helmholtz equation, we get \(\Delta G^{\circ} = T \Delta S^{\circ}-T \Delta S^{\circ}\). Simplifying this we get \(\Delta G^{\circ} = 0\). Thus, the value of \(\Delta G^{\circ}\) is zero.
03

Predict the value of \(K\)

The value of \(K\) can be predicted using the relationship between \(\Delta G^{\circ}\) and \(K\). According to the equation \(\Delta G^{\circ} = -RT \ln K\), where R is the universal gas constant and T is the absolute temperature, when \(\Delta G^{\circ} = 0\), the formula can be rearranged to \(\ln K = -\Delta G^{\circ}/RT\). Since we found that \(\Delta G^{\circ} = 0\) in the previous step, the value of \(K\) will be 1 as \(\ln1 = 0\).

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Most popular questions from this chapter

Eor a chemical reaction with \(K>1\), is \(\Delta G^{\circ}\) positive or negative?

a) What values of \(\Delta H^{\circ}\) indicate that a reaction is energetically favorable? b) What values of \(\Delta S^{\circ}\) indicate that a reaction is entropically favorable?

For a chemical reaction with \(K<1\), is \(\Delta G^{\circ}\) positive or negative?
See all solutions

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