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Chemist studying the following equilibirum; which has the given equilibrium constant at certain temperature:Nx(g) 2H,O(g) = 2NO(g) 2Hx(g)Kp = 3.He fills reaction ve...

Question

Chemist studying the following equilibirum; which has the given equilibrium constant at certain temperature:Nx(g) 2H,O(g) = 2NO(g) 2Hx(g)Kp = 3.He fills reaction vessel at this temperature with 15 atm of nitrogen gas and 35 atm of water vapor: Use this data to answer the questions in the table below:Can you predict the equilibrium pressure of Hz using only the tools yes available to You within ALEKS? 0 noDIf You sald yes_ then enter the equilibrium pressure of Hz at right: Round your answer to s

chemist studying the following equilibirum; which has the given equilibrium constant at certain temperature: Nx(g) 2H,O(g) = 2NO(g) 2Hx(g) Kp = 3. He fills reaction vessel at this temperature with 15 atm of nitrogen gas and 35 atm of water vapor: Use this data to answer the questions in the table below: Can you predict the equilibrium pressure of Hz using only the tools yes available to You within ALEKS? 0 no D If You sald yes_ then enter the equilibrium pressure of Hz at right: Round your answer to significant digit: atm



Answers

At $500^{\circ} \mathrm{C}, K$ for the formation of ammonia from nitrogen and hydrogen gases is $1.5 \times 10^{-5}$. $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Calculate the equilibrium partial pressure of hydrogen if the equilibrium partial pressures of ammonia and nitrogen are $0.015 \mathrm{~atm}$ and $1.2 \mathrm{~atm}$, respectively. 35. At a certain temperature, $K$ is 4.9 for the formation of one mole of bromine chloride gas (BrCl) from its elements. A mixture at equilibrium at this temperature contains all three gases. The partial pressures at equilibrium of bromine and chlorine gas is $0.19 \mathrm{~atm}$. What is the partial pressure of bromine chloride in this mixture at equilibrium?

First we will write the equilibrium constant expression K p. It's going to be equal to the pressure of the product. N O C L squared because of the coefficient divided by the pressure of Eno, also squared And then times the pressure of Cl two. And then we can plug in the values to calculate the value of the constant. So um NOCL is 1.2 atmospheres, So it's 1.2 squared. We omit the units for the K expression you have, N O is Um 0.05. Okay. And we'll square that. Yeah. And then cl two is 0.3. So then we just have to calculate 1.2 Squared, Divided by .05 Squared And then divided by .3 Is 2, 2 significant figures, 1.9 Times 10 to the 3rd. Okay. And that is the value of K p.

For part A With guts, huh? Psychlo Crew uh Psychlo Propane equilibrium with proteins Cape Easy Pull the K, C O R T, Delta and Casey is 1.0 times 10 to the negative. Five. Ours 0.8206 Temperature is 500 Calvin and moles of gas. There's one on each side, one minus one is zero. So adult and zero there for the KP is exactly the same. It's KC 1.0 times 10 to the negative five for B, the K P single with partial pressure of the protein over the partial pressure of the cyclo propane substitute or values in 1.0 times 10 to the negative five is equal Thio by 0.0. Partial pressure of the cyclo propane they're for the partial pressure of the cycle of propane is going to be equal to you 5.0 times 10 to the negative five atmospheres For part C question us. Can you alter the ratio of the concentrations by adding cycle of propane? So adding cycle of propane would cause a shift towards the rights, which means that overall, the concentrations of the two would increase the concentration of the pro peen would go up and the concentration of the cyclo propane would go down so we can alter the ratio by adding Psychlo propane bye, decreasing the volume decreasing volume that's also the seem as increasing the pressure. And since there's one mole of gas and inside, there's no shift that is going to take place with prospective licious highest principle. So it is not possible to alter the concentrations by affecting the volume for D, which has a larger re constants. And so K P is equal to K C, which is equal to K F over K R. And this is equal to 1.0 times 10 to the negative five, which is less than one. And, um, it's insist Rachel is less than one. Then the, um, the reverse kor is greater. The K f and for E y a cycle a propane so reactive, uh, say Chlo, propane is very reactive. I do too. Be high, uh, strain of the the ring or the cyclic compound. So when it's in, that cyclic configuration is

So here we have a specific reaction where we form ammonia leads to the formation of nitrogen and gases hydrogen. So we have all gaseous species Initially we essentially have 2.50 moller ammonia. And at equilibrium we see that we have essentially 0.10 moller of nitrogen, which means that the amount of nitrogen increased by the following three times as much hydrogen is produced as nitrogen which would result in the final concentration of hydrogen. And there's essentially two times as much ammonia that's consumed for nitrogen. And we have the following concentrations. So the equilibrium constant with respect to concentration, which we can essentially solve by writing our equilibrium constant, adjusting for psychometric coefficients. Okay. And we essentially see That the value is essentially about 5.1 times 10 to the -4. So it's important to remember that pressure can be expressed as the following by rearranging the ideal gas equation. While this gives essentially the concentration and since we essentially have, we essentially can add all the concentrations together uh and basically multiply by RT since basically we're assuming that this is an ideal gas and by doing so we can find the total pressure. Yeah. Okay. Mhm. Yeah. And we're assuming here that our temperature is 473 calvins. And using this information, we can find that our total pressure in this case is equivalent to about 105 atmospheres. And in our last part we're asked to find the equilibrium constant with respect to pressure which is equivalent to the equilibrium constant with respect to concentration multiplied by our our T to the power of the change of number of moles of gas. So we see that we have a change of number of moles of gas of two moles. So we can use this information to solve. Okay, Yeah. And using this information to solve, We find that our equilibrium constant with respective pressure is essentially about 0.77. And this gives our final answer.

So we want to know all the partial pressure values at equilibrium for all the products and reacting in this balanced chemical equation. So given the information that we have in the problem itself, we want to know partial pressure at equilibrium. We can start out by making an ice table which is initial change that at equal Aram So we were told nitrogen gas We have 0.862 as an initial value here but then 0.373 has their initial value here and we have no products initially. So we can dio minus x here minus three x here, this three here for the change and then a positive two x here an equilibrium We have zero point eight six to minus x 0.373 minus three x and then two x here. So since we were given the k equal a room constant value, we're able to put that here and make an equilibrium constant equation so that we can solve for that X. So we do our products raised the two of this coefficient here over reactive point 862 minus x multiplying by 0.37 three minus three hunger. So if you are to simplify this equation and solve it, you would find that X is equivalent to zero point 00 to a T. I am so now that we know what the X value is, we can go ahead and find the partial pressure values at equilibrium of nitrogen hydrogen. And our product is now we know all of the values we hold assault for this. Find the personal pressure of her first reacted we do 3.862 witches. Our initial minus X. We found to be 0.2 and this is cool it too. Zero point 86 Partial pressure of hydrogen is zero point 373 Linus three times our x excuse us of value of zero point 367 and then for our pressure of our product, we have two times are X value yet 0.4 So just that we could be sure that we have the correct partial pressures we found. Partial pressure of nitrogen is equivalent to zero point 86 18 AM because we carried those units throughout pressure of hydrogen 367 me too. And last but not least, our last partial pressure is 0.4 a. T and


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