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Given the following pair of equations:

$$\Delta { G }^{ \circ }=–RT \ln K\quad \text{and} \quad \Delta { G }^{ \circ }=\Delta { H }^{ \circ }-T\Delta { S }^{ 0 }$$

...it follows that:

$$–RT \ln K=\Delta { H }^{ \circ }-T\Delta { S }^{ 0 }$$

...which becomes:

$$\ln K=–\frac { { \Delta H }^{ \circ } }{ RT } +\frac { T\Delta { S }^{ \circ } }{ RT }$$

..and, finally:

$$\ln K=–\frac { \Delta { H }^{ \circ } }{ R } \left( \frac { 1 }{ T } \right) +\frac { \Delta { S }^{ \circ } }{ R }$$

Note that this is an equation for a straight line with $\ln K$ as the dependent variable ("y") and $\left( \frac { 1 }{ T } \right)$ as the independent variable ("x"). The slope is $(–\frac { \Delta { H }^{ \circ } }{ R } )$, and the intercept is $\frac { \Delta { S }^{ \circ } }{ R }$.

Consider the following table, showing the value of $\ln K$ as a function of the reciprocal of the Kelvin temperature for a hypothetical reaction:

$$\text{Reactants} \rightarrow \text{Products}$$

$\frac { 1 }{ T }$ $\ln K$
0.00050 25.0
0.00100 20.0
0.00150 15.0
0.00200 10.0
0.00250 5.00

$\$

A plot of the data looks like this.

Consider the following four statements concerning this system.

Which of the following does NOT correctly fit the system described here?

A

As temperature increases, the value of $K$ increases, so the reaction must be endothermic as written.

B

The value of $\cfrac { \Delta { H }^{ \circ } }{ R }$ is independent of temperature.

C

As the temperature of the system is increased, the position of equilibrium shifts to the right, favoring products.

D

As temperature increases, the value of $K$ decreases, so the reaction must be exothermic as written.

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