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?