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Thermodynamics and The Temperature Dependence of K

APCHEM-IYJLZS

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.

Created for Albert.io. Copyright 2016. All rights reserved.

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.