# 0.18 Reaction equilibrium in the gas phase  (Page 5/6)

 Page 5 / 6

## Observation 3: temperature dependence of the reaction equilibrium

We have previously observed that phase equilibrium, and in particular vapor pressure, depend on thetemperature, but we have not yet studied the variation of reaction equilibrium with temperature. We focus our initial study on this reaction and we measure the equilibrium partial pressures at a variety of temperatures. From these measurements, we can compile the datashowing the temperature dependence of the equilibrium constant ${K}_{p}$ for this reaction in [link] .

Equilibrium constant for the synthesis of hi
T (K) ${K}_{p}$
500 6.25 × 10 –3
550 8.81 × 10 –3
650 1.49 × 10 –2
700 1.84 × 10 –2
720 1.98 × 10 –2

Note that the equilibrium constant increases dramatically with temperature. As a result, at equilibrium, thepressure of HImust also increase dramatically as the temperature is increased.

These data do not seem to have a simple relationship between ${K}_{p}$ and temperature. We must appeal to arguments based on Thermodynamics, from which it is possible to show that theequilibrium constant should vary with temperature according to the following equation:

$\ln {K}_{p}=-\left(\frac{\Delta ({H}^{°})}{RT}\right)+\frac{\Delta ({S}^{°})}{R}$

If $\Delta ({H}^{°})$ and $\Delta ({S}^{°})$ do not depend strongly on the temperature, then this equation would predict a simple straight line relationship between $\ln {K}_{p}$ and $\frac{1}{T}$ . In addition, the slope of this line should be $-\left(\frac{\Delta ({H}^{°})}{R}\right)$ . We test this possibility with the graph in [link] .

In fact, we do observe a straight line through the data. In this case, the line has a negative slope. Notecarefully that this means that ${K}_{p}$ is increasing with temperature. The negative slope via [link] means that $-\left(\frac{\Delta ({H}^{°})}{R}\right)$ must be negative, and indeed for this reaction in this temperature range, $\Delta ({H}^{°})=15.6\frac{\mathrm{kJ}}{\mathrm{mol}}$ . This value matches well with the slope of the line in [link] .

Given the validity of [link] in describing the temperature dependence of the equilibrium constant, we can also predict that anexothermic reaction with $\Delta ({H}^{°})< 0$ should have a positive slope in the graph of $\ln {K}_{p}$ versus $\frac{1}{T}$ , and thus the equilibrium constant should decrease with increasing temperature. A good example of an exothermic reaction is the synthesis of ammonia for which $\Delta ({H}^{°})=-99.2\frac{\mathrm{kJ}}{\mathrm{mol}}$ . Equilibrium constant data are given in [link] . Note that, as predicted, the equilibrium constant for this exothermic reaction decreases rapidlywith increasing temperature. The data from [link] is shown in [link] , clearly showing the contrast between the endothermic reaction and the exothermic reaction. Theslope of the graph is positive for the exothermic reaction and negative for the endothermic reaction. From [link] , this is a general result for all reactions.

Equilibrium constant for the synthesis of ammonia
T (K) ${K}_{p}$
250 7.8 × 10 8
298 6 × 10 5
350 2 × 10 3
400 36

## Observation 4: changes in equilibrium and le châtelier's principle

One of our goals at the outset was to determine whether it is possible to control the equilibrium whichoccurs during a gas reaction. We might want to force a reaction to produce as much of the products as possible. In the alternative, ifthere are unwanted by-products of a reaction, we might want conditions which minimize the product. We have observed that theamount of product varies with the quantities of initial materials and with changes in the temperature. Our goal is a systematicunderstanding of these variations.

#### Questions & Answers

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Giriraj
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Damian
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Professor
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Professor
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scanning tunneling microscope
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nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
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biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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