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The equilibrium constant $K_{\mathrm{c}}$ for the following reaction is 1.2 at $375^{\circ} \mathrm{C}$$$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$(a) What is the value of $K_{P}$ for this reaction?(b) What is the value of the equilibrium constant $K_{\mathrm{c}}$ for $2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) ?$(c) What is the value of $K_{\mathrm{c}}$ for $\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)$$\rightleftharpoons \mathrm{NH}_{3}(g) ?$(d) What are the values of $K_{P}$ for the reactions described in (b) and (c)?

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a) $2.35 \times 10^{3}$b) 0.83c) $=1.1$d) 0.021

Chemistry 102

Chapter 14

Chemical Equilibrium

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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.

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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.

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The equilibrium constant $…

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The equilibrium constant f…

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Write the equilibrium cons…

So for the following problem, we want an answer. Some questions about equilibrium, constants in both KP and Casey. So starting with the balance equation that I've pre written here, we were given the Casey value. But we want to find the KP value. So in order to find the KP value, what we're gonna want to dio issues the following equation that relates equilibrium, constant and rivers two partial pressure to the equilibrium, constant regards to concentration and were able to find the KP because all of our substances in this equation are in the gaseous form. So talk him out. What we have to plug into this equation. The temperature. We were given this 355 degrees Celsius. We want to convert that to Calvin. So doing the conversion, we find 648 degrees Calvin. So we also confined the number of moules. And we can simply do that by the moles of gases product and then subtract from that value the moles of gaseous react which gives us native tomb. So now we can put that into this equation. Wanna find KP in the problem? We were given that Casey is equivalent to 1.2 or it is the ideal gas constant between now you multiply this but the temperature we found and raise it to a power of native to. And if we are to calculate this out, we find that the KP it's equivalent to 4.24 times 10 to the negative four. So now we have the K C and the KP for this balance equation. But we want to look at this same equation, but written a few different ways, go down a little bit to have room to work. And next, we're gonna look at what happens if we reverse the balanced equal of room that we looked at in the first part of this problem. So we have the same exact species in this reaction. The whole thing is just reversed, and we know when we're looking for K C. We have the Casey for the not reversed form, which is when we did in the first part, we found that what we were given that Casey was 1.2. So in order to find it in the reverse direction, all you have to do is one over the Casey that we knew, and that causes zero point 83 So now we can also find KP value since using our new equilibrium constant Casey. So we plug it in just like we did in the first part. But we used our new equilibrium constant Casey here. Everything else is the same. Nothing else has changed. It's still at the same temperature. We still have the same conversion that we found, but this is now going to be a positive, too. Is now we have for Delta end four for the products minus two for the reactive. It's so if we calculate this out, we find that the new KP is 2.35 times 10 to the third, which makes sense. So if we go down a bit more so we can look at the last part of this problem, this is the same format as the first part of this problem. But we're dividing all on both sides by two. So looking at the equilibrium constant Casey for this balanced you call a room. We know that the normal Casey for the normal version that we saw in the first part of this equilibrium was 1.2 And since we went ahead and divided both sides by two, we can do the square root of our original equilibrium constant in order to find this new Casey and we find that to be one point one. So now that we have a new equilibrium constant Casey again, we can find the KP for this balance equation as well. Went to plug in 1.1 are ideal gas. Constant is a constant, so it is always the same. We haven't changed the temperature, but we will see a change in what it is raised because we have changed the moles again. So Delta end, you're looking at one minus three halfs plus 1/2 which gives us value of negative one we're gonna raises to a power of negative one. If we calculate this out, we find this to be 0.0 21

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