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Write the equilibrium constant expressions K for the following reactions, CHalg) H;o (g) =colg) 3 Hz (g)b) 2 NzHa (I) + 2 NOz (g) == 3 Nz (g) + 4 H,o ()NHANOs (s) =...

Question

Write the equilibrium constant expressions K for the following reactions, CHalg) H;o (g) =colg) 3 Hz (g)b) 2 NzHa (I) + 2 NOz (g) == 3 Nz (g) + 4 H,o ()NHANOs (s) =2 H,O (g) N,O (8)4 HCI (aq) MnOz (s) MnClz (aq) + 2 Hzo ' W+C (e)Ca;(POa)z (s) = 3 Ca2+ (aq) 2 PO;?- (aq)

Write the equilibrium constant expressions K for the following reactions, CHalg) H;o (g) =colg) 3 Hz (g) b) 2 NzHa (I) + 2 NOz (g) == 3 Nz (g) + 4 H,o () NHANOs (s) =2 H,O (g) N,O (8) 4 HCI (aq) MnOz (s) MnClz (aq) + 2 Hzo ' W+C (e) Ca;(POa)z (s) = 3 Ca2+ (aq) 2 PO;?- (aq)



Answers

Write equilibrium constant expressions $(K)$ for the following reactions: (a) $2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \rightleftharpoons$ $$ \begin{array}{l} \text { (b) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+3 \mathrm{Cu}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \\ \text { (c) } \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s) \end{array} $$

In this video we're going to write equilibrium constant expressions for the following reactions. So um let's see. We have a lot of species. So we'll write the Casey expressions. Remember that solids and liquids are not included. So we're going to look for us at the products for the top. We have um N. O. Concentration is squared because there's a coefficient of two And then see you two plus. And we'll put a cube corresponding to the coefficient and then the water is a liquid. So we omit it and on the bottom we'll have our reactions, We ignore the solid here. So we'll have um just to you know, three minus and that's gonna be squared. And then age plus Raised to the 8th power for part B will omit the solid on both sides. And we're going to look first at the product, we have just one relevant product. So to that's going to be squared. And then on the bottom we have the reactant 02 and that's going to be raised to the third power, and then for part C. So we only have a solid product, nothing else. So that means we'll just put a one on the top to stand in, and then we have See a two plus And CO three two minus no exponents, because all the coefficients are one, and we're done.

Okay. So for each of these we're gonna, right Casey so Casey is products Overreactions raised to their coefficients? Mhm. So one of our products is no two. So the concentration of N. 02 times the concentration of N. O. Those are our products. And then we'll divide it by concentration of our reactant and 203. So that's your expression for that reaction. The second one We've got the concentration of H. two And we square it because there's a coefficient of two Times concentration of S. two And we'll divide that by the concentration of H two. S again squared because there's a coefficient of two. And the 3rd 1 We got to go to. So the concentration of co two squared divided by the concentration of N. O. Squared Times The Concentration of Co two. And then finally For the final one we have the concentration of p NH 23 times the concentration of hcl cubed. And then we'll divide that by the concentration of our PCL three and times are concentration of RNH three. Again we need a cube that

Okay. So we're going to write some expressions for K here. Okay. And the general format is products overreacted. It's raised to their coefficients and we're going to ignore liquids and solids. They don't get included. Okay, So for this first one, we have two of them are solid, so we're going to go ahead and ignore them. So in this case the Okay, simply equals the partial pressure of CO two. Okay, was for the second one we've got a gas of liquid and a couple of gases. So we are going to ignore the liquid. All right. It's okay. Is going to equal the partial pressure of ceo squared and that's what's coefficient times of partial pressure of hydrogen to the fifth. And then we're going to divide that all by the partial pressure of C two H six and we'll ignore the water. Mm for C. Looks like we've got all gases. So again, products overreacted. It's so the partial pressure of the NH three for the fourth in terms of partial pressure of CO. Two to the fifth, provided by the partial pressure of R. N. O. To the fourth, and our partial pressure of our water vapor to the sixth. And then our last one, we've got a gas, a liquid and a solid. So we're only going to worry about the gas, which is a reactant. It's okay, ends up being one divided by the partial pressure of NH three.

In this video we are going to write equilibrium constant expressions for the following equations. Remember that we omit um solids and liquids from the expression we assume that the pressure and concentration won't change much. So for part A We have the equilibrium constant is equal to the pressure of Co two which is a product Divided by uh nothing because there is there solid. So it's actually just the pressure of co two for part B. We have the pressure of ceo and that's going to be squared because of the coefficient Times the pressure of H2 and that's going to be raised to the 5th power divided by the pressure of C two. H 6. For part C. We have the pressure of NH three races to the 4th power Time side of 02 raised to the 5th power Divided by the pressure of a no to the 4th Times the pressure of H 20 to the 6th. And lastly for part D. We have no gosh mr Aquarius products so put one over and then on the bottom we have the pressure of and H three. And routine


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