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At a certain temperature the following reactions have the constants shown:$$\begin{array}{cc}\mathrm{S}(s)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g) & K_{\mathrm{c}}^{\prime}=4.2 \times 10^{52} \\2 \mathrm{S}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) & K_{c}^{\prime \prime}=9.8 \times 10^{128}\end{array}$$Calculate the equilibrium constant $K_{\mathrm{c}}$ for the following reaction at that temperature:$$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$

$$K_{c}=5.6 \times 10^{23}$$

Chemistry 102

Chapter 14

Chemical Equilibrium

Carleton College

University of Maryland - University College

University of Toronto

Lectures

10:03

In thermodynamics, a state of thermodynamic equilibrium is a state in which a system is in thermal equilibrium with its surroundings. A system in thermodynamic equilibrium is in thermal equilibrium, mechanical equilibrium, electrical equilibrium, and chemical equilibrium. A system is in equilibrium when it is in thermal equilibrium with its surroundings.

00:54

In chemistry, chemical equilibrium (also known as dynamic equilibrium) is a state of chemical stability in which the concentrations of the chemical substances do not change in the course of time due to their reaction with each other in a closed system. Chemical equilibrium is an example of dynamic equilibrium, a thermodynamic concept.

02:15

At a given temperature, th…

06:18

At a certain temperature, …

01:04

The equilibrium constant f…

03:16

02:48

01:09

For each reaction, an equi…

00:43

So in this problem, we want to find the Casey for this balance chemical equation and where you're given the following two equilibrium constants being with. So in order to find what the Casey that were missing for this one is we're gonna first look at fact that we know that, IHS Plus we'll to in the gases form So a sulfur in an auction react to create. So for dioxide, we know that this Casey is equivalent to 4.2 times 10 to the 52nd. But because the S or two that we need is on this side, we need to reverse this. And we also need to multiply by Choose that we have the same reactive with the same coefficients. We'll carry that and multiply my two throughout. Carry that through. So that means for the Casey this Casey, we will raise it to the negative to. And we're doing this because we reversed it, which is why there's a negative. So that means now, Casey value for this problem or for this equation is to the negative to this is a 52. So now Bill need the second equation were given to s some would. So two sulphur waas 302 gaseous form react to make to s 03 gas use for And we know that this Casey it's equivalent to 9.8 times 10 to the 100 in 28. And since we didn't change this equation at all, we do not have to change the second given equilibrium constant. But we are not looking at this one since we have changed it. So that means going to look at what we have now this in this cancel. So we have to eso to pause Oh too to s 03 which is what we need that we've looking for during this problem. So that means this case see, value is equivalent to the first colistin that were given here, raised to the negative two, which is what we did here, multiplied by the second equal of rebel constant that we didn't end up changing in this. So if we calculate what this Casey is going to be, we see that it is 4.2 times 10 to the 52nd raised the negative, too. Times 9.8 has 10 the 128 which, when calculated through, comes out to be our answer of 5.6 times 10 to the 23

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