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Math  /  Data & Statistics

QuestionThe statements in the tables below are about two different chemical equilibria. The symbols have their usual meaning, for example ΔG∘\Delta G^{\circ} stands for the standard Gibbs free energy of reaction and KK stands for the equilibrium constant.
In each table, there may be one statement that is false because it contradicts the other three statements. If you find a false statement, check the box next to it. Otherwise, check the "no false statements" box under the table. \begin{tabular}{|r|c|} \hline statement & false? \\ \hline ln⁡K=−1\ln K=-1 & 0 \\ \hlineK=1K=1 & 0 \\ \hlineΔS∘=ΔH∘T\Delta S^{\circ}=\frac{\Delta H^{\circ}}{T} & \\ \hlineΔG∘=0\Delta G^{\circ}=0 & \\ \hline \end{tabular} no false statements: \begin{tabular}{|r|c|} \hline statement & false? \\ \hlineΔG∘>0\Delta G^{\circ}>0 & 0 \\ \hline ln⁡K<0\ln K<0 & 0 \\ \hlineΔS∘<ΔH∘T\Delta S^{\circ}<\frac{\Delta H^{\circ}}{T} & 0 \\ \hlineK>1K>1 & 0 \\ \hline \end{tabular} no false statements:

Studdy Solution

STEP 1

What is this asking? We need to find the lie in each table of thermodynamic statements about chemical reactions at equilibrium! Watch out! Don't forget those important thermodynamic relationships!

STEP 2

1. Analyze the first table
2. Analyze the second table

STEP 3

The first statement says ln⁡K=−1\ln K = -1.
If this is true, then K=e−1K = e^{-1}, which is definitely *not* equal to 11.
So, the second statement, K=1K=1, contradicts the first!

STEP 4

The fourth statement says ΔG∘=0\Delta G^{\circ} = 0.
We know the super important equation ΔG∘=−R⋅T⋅ln⁡K\Delta G^{\circ} = - R \cdot T \cdot \ln K.
If ΔG∘=0\Delta G^{\circ} = 0, then ln⁡K\ln K *must* also be zero, which means K=1K=1.
This agrees with the second statement but contradicts the first!

STEP 5

The third statement tells us ΔS∘=ΔH∘T\Delta S^{\circ} = \frac{\Delta H^{\circ}}{T}.
Remember the Gibbs free energy equation: ΔG∘=ΔH∘−T⋅ΔS∘\Delta G^{\circ} = \Delta H^{\circ} - T \cdot \Delta S^{\circ}.
Substituting the third statement into this equation gives us ΔG∘=ΔH∘−T⋅ΔH∘T=ΔH∘−ΔH∘=0\Delta G^{\circ} = \Delta H^{\circ} - T \cdot \frac{\Delta H^{\circ}}{T} = \Delta H^{\circ} - \Delta H^{\circ} = 0.
This agrees with the fourth statement!

STEP 6

Since statements 2, 3, and 4 all agree with each other, and statement 1 contradicts them, statement 1, ln⁡K=−1\ln K = -1, must be the false one!

STEP 7

The first statement claims ΔG∘>0\Delta G^{\circ} > 0.
We know that ΔG∘=−R⋅T⋅ln⁡K\Delta G^{\circ} = - R \cdot T \cdot \ln K.
Since RR and TT are always positive, if ΔG∘\Delta G^{\circ} is positive, then ln⁡K\ln K *must* be negative.
This agrees with the second statement, ln⁡K<0\ln K < 0!

STEP 8

If ln⁡K<0\ln K < 0, then KK must be between 00 and 11.
This directly contradicts the fourth statement, which says K>1K > 1!

STEP 9

The third statement says ΔS∘<ΔH∘T\Delta S^{\circ} < \frac{\Delta H^{\circ}}{T}.
Multiplying both sides by −T-T (and flipping the inequality sign since −T-T is negative), we get −T⋅ΔS∘>−ΔH∘-T \cdot \Delta S^{\circ} > -\Delta H^{\circ}.
Adding ΔH∘\Delta H^{\circ} to both sides gives us ΔH∘−T⋅ΔS∘>0\Delta H^{\circ} - T \cdot \Delta S^{\circ} > 0.
Since ΔG∘=ΔH∘−T⋅ΔS∘\Delta G^{\circ} = \Delta H^{\circ} - T \cdot \Delta S^{\circ}, this means ΔG∘>0\Delta G^{\circ} > 0, which agrees with the first statement!

STEP 10

Since statements 1, 2, and 3 agree, and statement 4 contradicts them, statement 4, K>1K > 1, must be false!

STEP 11

In the first table, the false statement is ln⁡K=−1\ln K = -1. In the second table, the false statement is K>1K > 1.

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