One step in the commercial synthesis of sulfuric acid is the...

One step in the commercial synthesis of sulfuric acid is the catalytic oxidation of sulfur dioxide: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ (a) A mixture of 192 $\mathrm{g}$ of $\mathrm{SO}_{2}, 48.0 \mathrm{g}$ of $\mathrm{O}_{2}$ and a $\mathrm{V}_{2} \mathrm{O}_{5}$ catalyst is heated to 800 $\mathrm{K}$ in a 15.0 $\mathrm{L}$ vessel. Use the data in Appendix $\mathrm{B}$ to calculate the partial pressures of $\mathrm{SO}_{3}$ $\mathrm{SO}_{2},$ and $\mathrm{O}_{2}$ at equilibrium. Assume that $\Delta H^{\circ}$ and $\Delta S^{\circ}$ are independent of temperature. (b) Does the percent yield of $\mathrm{SO}_{3}$ increase or decrease on raising the temperature from 800 $\mathrm{K}$ to 1000 $\mathrm{K}$ ? Explain. (c) Does the total pressure increase or decrease on raising the temperature from 800 $\mathrm{K}$ to 1000 $\mathrm{K}$ ? Calculate the total pressure (in atm) at 1000 $\mathrm{K}$

One step in the commercial synthesis of sulfuric acid is the catalytic oxidation of sulfur dioxide:
$$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$
(a) A mixture of 192 $\mathrm{g}$ of $\mathrm{SO}_{2}, 48.0 \mathrm{g}$ of $\mathrm{O}_{2}$ and a $\mathrm{V}_{2} \mathrm{O}_{5}$ catalyst is heated to 800 $\mathrm{K}$ in a 15.0 $\mathrm{L}$ vessel. Use the data in Appendix $\mathrm{B}$ to calculate the partial pressures of $\mathrm{SO}_{3}$ $\mathrm{SO}_{2},$ and $\mathrm{O}_{2}$ at equilibrium. Assume that $\Delta H^{\circ}$ and $\Delta S^{\circ}$ are independent of temperature.
(b) Does the percent yield of $\mathrm{SO}_{3}$ increase or decrease on raising the temperature from 800 $\mathrm{K}$ to 1000 $\mathrm{K}$ ? Explain.
(c) Does the total pressure increase or decrease on raising the temperature from 800 $\mathrm{K}$ to 1000 $\mathrm{K}$ ? Calculate the total pressure (in atm) at 1000 $\mathrm{K}$

Key Concepts

Ideal Gas Law

The ideal gas law (PV = nRT) describes the relationship between pressure, volume, temperature, and the number of moles of a gas. It is a critical tool for calculating the pressures and amounts of gases in equilibrium, particularly in reactions involving gaseous species.

Le Chatelier's Principle

Le Chatelier's Principle states that if an external change is imposed on a system at equilibrium, the system will adjust to counteract that change and re-establish equilibrium. This principle is used to predict the direction in which the equilibrium will shift when variables such as temperature or pressure are altered.

Effect of Temperature on Equilibrium

The effect of temperature on equilibrium involves understanding that the equilibrium constant is temperature-dependent. Changes in temperature can favor either the exothermic or endothermic direction of the reaction, thereby altering the yield of products, even though the total number of moles or the overall reaction stoichiometry remains constant.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a balanced chemical equation. It is essential for determining how initial amounts of substances are converted into products as the reaction proceeds toward equilibrium, based on the mole ratios indicated by the balanced equation.

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible reaction where the forward and reverse reaction rates are equal, resulting in no net change in the concentrations or partial pressures of the reactants and products. This concept is fundamental for predicting the composition of the system under given conditions and is quantified by the equilibrium constant.

Partial Pressure

Partial pressure is the pressure that an individual gas in a mixture would exert if it occupied the entire volume by itself. In gas-phase equilibrium calculations, the partial pressures of the reacting species are used with the ideal gas law to relate the number of moles to the pressure, thereby allowing determination of the equilibrium composition.

Transcript

00:01 We are given this reaction, and in part a, we want to determine what the equilibrium partial pressure of so2 is given the values for the equilibrium partial pressures for o2 and so3.

00:12 We are told a value of kc for this reaction, but we need to convert this into kp, because when we form the equilibrium expression for kp, the concentrations of the gases are measured in terms of partial pressures.

00:26 And we already have the equilibrium concentrations in terms of partial pressures for two out of these three gases.

00:33 So if we can find the value of kp, we can use that to solve for the partial pressure of the remaining gas at equilibrium.

00:41 So we have a value for kc and we first need to convert it into kp.

00:46 We do that using the equation kp equals kc times rt to the power of delta n gas.

00:55 The given value of kc at 600 kelvin, and the problem is 1 .7 times 10 to the 8th.

01:09 We multiply this by r, the ideal gas constant, 0 .0 821 liters times atmospheres per mole times kelvin, times the given temperature of 600 kelvin.

01:29 We raise this to the power of delta n gas.

01:32 We have two total moles of gas on the products and two plus one, so three total moles of gas on the reactant side.

01:39 2 minus 3 is negative 1, so delta n gas is negative 1.

01:44 When we solve for kp now, we should get a value of 3 .45 times 10 to the 6th.

01:55 Now we can form that equilibrium expression for kp.

01:58 Based on the reaction.

02:00 We know that kp is equal to the partial pressure of so3 at equilibrium squared divided by the partial pressure of so2 at equilibrium squared times the partial pressure of o2 at equilibrium.

02:25 And now we can rearrange this expression to solve for the partial pressure of so2 at equilibrium.

02:30 And we do that, we see that pso2 is equal to a square root of pso3 squared divided by kp times p .o2.

02:52 Now we can plug in the values that we are given in the problem to solve for pso2.

02:57 We're told that pso3 at equilibrium is 300 atmospheres.

03:02 And we square that.

03:05 Kp, we solved for to be 3 .45 times 10 to the sixth.

03:14 And p .o .2 at equilibrium, we were told, is 100 atmospheres.

03:21 So when we plug that all in, our final answer for the equilibrium partial pressure of s .o .2 at 600 kelvin is equal to 0 .016 .0 .6.

03:37 Atmospheres.

03:41 And now in part b we are given information about the initial amounts of both of the products in this reaction system.

03:47 And we want to determine from this information what the value of kc is at a new temperature of 1 ,000 kelvin.

03:54 We also want to know what the partial pressure of so2 is at this new temperature.

04:03 So we're going to need to set up a reaction table in order to help us determine that value for kc.

04:12 So the reaction again is 2 -2 gas plus o2 gas goes to 2 -s -o -3 gas.

04:31 We set up our reaction table with the initial concentrations, the change in the equilibrium expressions...

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