A reaction vessel at 27^∘ C contains a mixture of SO2(P=3.00 atm) and O2(P=1.00 atm ) . When a c

A reaction vessel at 27^∘ C contains a mixture of SO2(P=3.00...

A reaction vessel at $27^{\circ} \mathrm{C}$ contains a mixture of $\mathrm{SO}_{2}(P=3.00 \mathrm{atm})$ and $\mathrm{O}_{2}(P=1.00$ atm $) .$ When a catalyst is added, this reaction takes place: $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$ At equilibrium, the total pressure is 3.75 atm. Find the value of $K_{\mathrm{c}}.$

A reaction vessel at $27^{\circ} \mathrm{C}$ contains a mixture of $\mathrm{SO}_{2}(P=3.00 \mathrm{atm})$
and $\mathrm{O}_{2}(P=1.00$ atm $) .$ When a catalyst is added, this reaction takes
place: $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$
At equilibrium, the total pressure is 3.75 atm. Find the value of $K_{\mathrm{c}}.$

Key Concepts

Equilibrium Constant Expression

The equilibrium constant expression quantifies the ratio of product to reactant concentrations (or partial pressures) at equilibrium, each raised to the power of their stoichiometric coefficients. This expression, whether in terms of concentration (Kc) or pressure (Kp), is fundamental for predicting the position of equilibrium.

Ideal Gas Law in Equilibrium Calculations

The ideal gas law (PV = nRT) links pressure, volume, number of moles, and temperature in a gaseous system. In equilibrium problems, it is used to relate changes in pressure to changes in concentration or number of moles, thereby facilitating the conversion of measured pressures into molar concentrations.

Conversion between Kp and Kc

For gaseous reactions, the relationship between the equilibrium constant expressed in terms of partial pressures (Kp) and the one expressed in terms of concentrations (Kc) is given by the equation Kp = Kc(RT)^(?n), where ?n is the change in the number of moles of gas, R is the ideal gas constant, and T is the temperature. This conversion is essential when experimental conditions yield pressures but the equilibrium calculations require concentrations.

Chemical Equilibrium

Chemical equilibrium is the state in which the forward and reverse reactions occur at the same rate, resulting in constant concentrations of reactants and products over time. Understanding equilibrium is essential for determining how a reaction responds to changes in conditions and for calculating equilibrium constants.

Reaction Stoichiometry

Reaction stoichiometry involves using the balanced chemical equation to relate the amounts of reactants consumed and products formed. The coefficients in the equation determine the proportional changes in quantities, which are crucial for setting up and solving equilibrium problems involving gases.

Transcript

00:01 So in this question, we are told that we have a reaction between heso2 and o2 to produce so3.

00:08 And we are told that we are given three atmosphere and one atmosphere, heso2 and o2 respectively.

00:15 And the total pressure at equilibrium is 3 .75 atmosphere.

00:20 So they want us to calculate the kc at equilibrium, which is the equilibrium constant.

00:26 And that is for concentration.

00:30 So during reaction, assume that we use 2x of s .o2, then we have to use x of this whole 2.

00:40 Because 2 to 1 is this dichotometry and then you produce 2x of xo3.

00:45 So at equilibrium, you have to subtract or hard altogether.

00:49 So 3 minus 2 hex, 1 minus x and this is 2x.

00:54 So the total pressure will be the partial pressure of heso2, o2, hana, that will be pso2 plus o2 plus heso3.

01:06 And we are told that the total pressure is 3 .75 atmosphere, so this would be equal to the partial pressure of s02, which is 3 minus 2x, plus the partial pressure of o2, which is 1 minus x, plus the partial pressure of s .o .3, which is 2x.

01:21 So you sum this together, you arrive at 4 minus 0 ,000, plus the partial pressure of s.

01:25 So you serve for hex, x is 0 .25.

01:29 So if x is 0 .25, then we can calculate the partial pressure of s .o2, whole 2, and s .o3, by substituting 0 .25 as x.

01:38 So this will be 3 minus 2 multiplied by 0 .25.

01:42 So this will give us 2 .5, and this will give us 0 .75 for s .3...

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