At 1000 K, Kp = 2.1 * 106 and ΔH° = - 107.7 kJ for the reaction H...

Question

At 1000 K, Kp = 2.1 * 106 and ΔH° = - 107.7 kJ for the reaction H21g2 + Br21g2 ∆ 2 HBr1g2.
(b) For the equilibrium in part (a), each of the following changes will increase the equilibrium partial pressure of HBr. Choose the change that will cause the greatest increase in the pressure of HBr, and explain your choice.
(ii) Adding 0.10 mol of Br2

Accepted Answer

hi everyone for this problem. It reads the reaction nitrogen gas plus hydrogen gas yields ammonium has the following values at 298 kelvin A K p equal to 6.2 times 10 to the fifth and a standard entropy change equal to negative 92.2 kg jules. Consider the following changes below which will result in an increase in the partial pressure of ammonia at equilibrium. Which of the changes will result in the highest increase in the partial pressures of ammonia justify your answer. So the question that we want to answer here is which will result in the highest increase in the partial pressure of ammonia. Okay, so let's take a look at our reaction. Okay, so we have our reaction and we have the K. P value and then the problem we're told it's equal to 6. times 10 to the fifth. So this reaction essentially goes to completion And we're also given the standard entropy change negative 92.2 killer jewels. Which means that this is eggs a thermic. Okay, so our standard entropy change Is negative .2, which means this is eggs are thermic. Okay, so let's take a look at our answer choices. So lowering the temperature. Okay, so if we lower the temperature, what this means is the equilibrium is going to shift to the right, so we can write that here shifts Right now looking at our reaction. If the equilibrium shifts to the right, that means there's going to be an increase in the amount of product which is ammonia. So we can right here leads to an increase of ammonia. So that's what would happen if we lowered the temperature. How about adding two moles of nitrogen gas or two moles of hydrogen gas? So these are two react ints. What's going to happen when we add these two reactant? The first thing we're going to want to do is see which one is the limiting reactant. Okay, so let's go ahead and figure that out. So we have 0.2, we're adding 0.2 moles of nitrogen gas and we want to know how much ammonia is this nitrogen gas going to produce. And we're going to use the multiple ratio between nitrogen gas and ammonia to see what this value is going to be. Okay, so looking at our balanced equation for every one mole of nitrogen gas, there is two moles Of Ammonia. So our moles of nitrogen gas cancel and we're left with moles of ammonia. So essentially the 0.2 moles of nitrogen gas is going to give 0.4 moles of ammonia. So let's go ahead and do the same thing for a hydrogen gas. So we're adding 0.2 moles of hydrogen gas. Okay, so we want to know how much moles of Ammonia is this hydrogen gas going to produce. So we're going to use the multiple ratio here as well. And For every three moles of hydrogen gas there are two moles Of ammonia. Okay, so our units for moles of hydrogen gas cancel. And we can figure out our moles of product. So we get 0. moles of ammonia. So here we can see that the hydrogen gas is the limiting reactant and it's the limiting reactant because it produces the smallest the smaller amount of the product. Okay, so based off of our answer choices, we want to know which of the changes will result in the highest increase in the partial pressure of ammonia. Okay, so adding two moles of hydrogen will result in the highest increase in partial pressure of ammonia because it's the limiting reactant. So that's going to make answer choice B number B or Yes. This one. The second one. This is going to be the correct answer. Okay, so that is it for this problem. I hope this was helpful.

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