At 900 C, Kc = 0.0108 for the reaction
CaCO31s2 Δ CaO1s2 + CO21g2...

Question

At 900 C, Kc = 0.0108 for the reaction
CaCO31s2 Δ CaO1s2 + CO21g2
A mixture of CaCO3, CaO, and CO2 is placed in a 10.0-L
vessel at 900 C. For the following mixtures, will the
amount of CaCO3 increase, decrease, or remain the same
as the system approaches equilibrium?
(c) 30.5 g CaCO3, 25.5 g CaO, and 6.48 g CO2

Accepted Answer

Hi everyone for this problem. It reads the equilibrium constant for the reaction at 820°C is 1.02 and were given the following reaction rubidium chlorate yields rubidium oxide plus chlorine dioxide in a 5.60 liter vessel at 820 degrees Celsius, A mixture of 67.2 g of rubidium chlorate, 23.4 g of rubidium oxide and 7. g of chlorine dioxide was placed as the system approaches equilibrium. What happens to the amount of rubidium chlorate? So the question that we're trying to answer in this problem is what happens to the amount of rubidium chlorate and an important piece of information we're given in this problem is the value for our equilibrium constant and that value is 1.02. So in order for us to know what happens to the amount of rubidium chlorate. We're going to need to know what is the value of our reaction quotient Q. When we figure out that value will be able to compare K two Q. To see which direction our equilibrium will go. Okay, so when K is greater than Q. That means the forward direction is going to be favored to reach equilibrium. When K is less than Q. That means the reverse direction is going to be favored to establish equilibrium and when K is equal to Q. That means the system is at equilibrium. Okay, so our goal here is to compare K two Q. We were given K in the problem. Que our equilibrium constant is equal to 1.02. Now, if we write this equilibrium expression, we're going to get the concentration of products over our concentration of reactant. But looking at our reaction something to remember is that pure solids and pure liquids are not included in the equilibrium expression. So we can go ahead and cross out our solids here. So that means our equilibrium expression is only going to be our concentration of chlorine dioxide, our concentration of products over our reactant, but we don't include pure solids. So this is our equilibrium expression. So our goal here is to find out what is our value for Q. But the thing is these expressions are the same. It's the values that are different. Okay, so for our K. This is going to be our equilibrium concentrations and Q is our non equilibrium concentrations. So our goal here is to find out our non equilibrium for chlorine dioxide. And remember concentration is in moles over leader. So our goal here is to find our concentration of chlorine dioxide. So we can find our value for Q. Okay, so let's go ahead and do that. So we know we need moles over leader in the problem, we are given our mass of chlorine dioxide. We're told it is let's do it in a different color. We're told our mass is 7.89 g of chlorine dioxide. So our goal is to go from grams to moles because we want concentration. Okay, so we'll use our molar mass for chlorine dioxide to go from grams to moles. So we'll use our periodic table for this. So in one mole of Chlorine dioxide, the molar mass is 67.45g. Okay, So here we see that our units for g cancel and we're left with moles. So when we do this calculation we get 0.117 moles of chlorine dioxide. So we have the numerator, we have moles but we need volume because concentration is an moles over leader. And the problem we're told are vessel is 5.60 L. So we'll go ahead and divide this by 5.60 liters. And when we do that we get a concentration equal to 0.373 molar. And remember mole is is moles over leader. Its concentration. Okay, so now that we know are Value for Q. We can compare Q 2K. Okay. So we said K is 1.02. Okay. And we now know Q is 0.00373. Okay. 00373. So based off of this we see that K is greater than Q. Okay. And so what we said that means is since K is greater than Q. Our reaction is going to shift in the four direction to establish equilibrium. So looking at our reaction if we're shifting to in the forward direction, we're shifting to the right, which means our concentration of reactant is going to decrease and our concentration of product is going to increase. Okay? So in terms of our rubidium chlorate, this means the amount is going to decrease. Okay, and that is going to be the answer for this problem. Our amount of rubidium chlorate decreases. That's the end of this problem. I hope this was helpful.

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