Consider the reaction: 2 NO(g)+O2(g) ⇌2 NO2(g) The following data show the equilibrium constant

Consider the reaction: 2 NO(g)+O2(g) ⇌2 NO2(g) The following...

Consider the reaction: $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ The following data show the equilibrium constant for this reaction measured at several different temperatures. Use the data to find $\Delta H_{\mathrm{rxn}}^{\circ}$ and $\Delta S_{\mathrm{rxn}}^{\circ}$ for the reaction. $$ \begin{array}{cc} \text { Temperature } & \kappa_{\mathrm{p}} \\ \hline 170 \mathrm{~K} & 3.8 \times 10^{-3} \\ \hline 180 \mathrm{~K} & 0.34 \\ \hline 190 \mathrm{~K} & 18.4 \\ \hline 200 \mathrm{~K} & 681 \\ \hline \end{array} $$

Consider the reaction:
$$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$
The following data show the equilibrium constant for this reaction measured at several different temperatures. Use the data to find $\Delta H_{\mathrm{rxn}}^{\circ}$ and $\Delta S_{\mathrm{rxn}}^{\circ}$ for the reaction.
$$
\begin{array}{cc}
\text { Temperature } & \kappa_{\mathrm{p}} \\
\hline 170 \mathrm{~K} & 3.8 \times 10^{-3} \\
\hline 180 \mathrm{~K} & 0.34 \\
\hline 190 \mathrm{~K} & 18.4 \\
\hline 200 \mathrm{~K} & 681 \\
\hline
\end{array}
$$

Key Concepts

Equilibrium Constant

The equilibrium constant is a dimensionless value that quantifies the ratio of the concentrations or partial pressures of products to reactants at equilibrium. Its value is temperature dependent and provides insight into the position of equilibrium for a chemical reaction.

Gibbs Free Energy Relationship

The relationship between the Gibbs free energy change and the equilibrium constant, expressed as ?G° = -RT ln K, connects thermodynamics and equilibrium. Here, ?G° is the standard change in free energy, R is the gas constant, T is the temperature, and K is the equilibrium constant. This equation is fundamental in understanding how changes in conditions affect the spontaneity of a reaction.

van't Hoff Equation

The van't Hoff equation, given by ln K = -?H°/(RT) + ?S°/R, relates the natural logarithm of the equilibrium constant to the inverse of the temperature. It provides a method to determine the standard reaction enthalpy (?H°) and entropy (?S°) by analyzing how the equilibrium constant changes with temperature, typically through a linear plot of ln K versus 1/T.

Standard Reaction Enthalpy (?H°)

Standard reaction enthalpy represents the heat absorbed or released during a chemical reaction under standard conditions. In the context of the van't Hoff analysis, it is determined from the slope (-?H°/R) of the ln K vs. 1/T plot, indicating whether a reaction is exothermic or endothermic.

Standard Reaction Entropy (?S°)

Standard reaction entropy reflects the change in disorder or randomness when a reaction proceeds under standard conditions. It can be extracted from the intercept (?S°/R) of the ln K vs. 1/T plot. This concept is crucial for understanding the overall thermodynamic favorability of a reaction when combined with the enthalpy change.

Transcript

00:01 We have been provided a reaction and for this reaction we have been asked the value of delta h reaction and the value of delta s not reaction.

00:11 These are the equation and we have been further provided a table of temperature versus equilibrium constant for the given reaction.

00:19 So we know a basic relation that equilibrium constant l m k, here k is the equilibrium constant is equals to minus delta h0 reaction divided by universal gas constant 1 upon temperature and kelvin plus delta s reaction divided by universal gas constant.

00:46 Okay.

00:48 So if we compare this with a line straight line whose equation is y equals to mx plus c we can see that m is the slope and this slope will be equal to minus delta h reaction over r okay and we are comparing y with l nk and x with one upon t and if we calculate the slope we can get the value of delta h of the reaction while this is the value of intercept okay this is the value of intercept so our main task is to find a equation between one upon t versus lm k so, how we can find this? we have different value of temperature versus equilibrium constant.

01:35 So if we calculate 1 upon t versus lnk values.

01:40 So for our first case, our 1 upon t value will be 0 .00582.

01:48 Our lnk at this temperature is minus 5 .5727.

01:54 Okay.

01:55 Now second, for second temperature, 1 upon t we just need to to the reciprocal will be 0 .00556 and the value of lnk, that is the equilibrium constant to the log, is equal to minus 1 .07 -881.

02:18 So as we know that this graph will form a straight line.

02:22 So we know that how to find the equation of a straight line when two points are given.

02:29 So equation of any straight line, when two of its coordinates are given, is given by y minus y1 is equals to y2 minus y1 over x2 minus x1.

02:42 X minus x1.

02:44 So our y, as we can see, y1 is the coordinate.

02:51 We can this, see, this is lnk versus 1 upon t.

02:57 So this will be y and this will be x.

02:59 So, y1 is over minus 5 .5727...