Calculating Kp for Decomposition of NH4Cl at High Temperature

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In summary, at equalibrium the partial pressure of NH3 (g) is 1.346 atm. The equilibrium constant for the decomposition of NH4Cl is 1.817*10-8*(8.3144*1197.4K).
  • #1
Masschaos
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A 9g sample of solid NH4Cl is heated in a 4L container to 924.4oC and decomposes according to the following equation.
NH4Cl(s) -> HCl(g) + NH3(g)

at equalibrium the partial pressure of NH3 (g) is 1.346 atm. Calculate the equilibrium constant for Kp for this reaction


Homework Equations



Kp = Pproductn/Preactantn

or

Kc equation and Pv = nRT

The Attempt at a Solution



First I used the Ideal gas equation the amount of NH3 in the system.
So n = PV/RT
n = (136383.45 Pa * (4 *10-3 L))/(8.3144*1197.4K)
n = 5.48*10-5

This means the concentration of the NH4Cl is.
n = 9/54.45 = 0.1653
0.1653 - 2(5.48*10-5) = 0.1652

I assumed then that Cl would also have this concentration.
So the Kc of the reaction = ([5.48*10-5*[5.48*10-5)/0.1652 = 1.817 * 10-8

Then Kp = Kc(RT)(delta)n
Kp= 1.817 * 10-8 * (8.3144*1197.4K)1
Kp = 0.00018 = 1.8*10-4

The more I look at this the more I feel there is a much easier solution, is my solution even right?
 
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  • #2
You have to calculate Kp, you are given pressure, why do you calculate concentrations instead of using pressure directly?

Besides, what is the activity of the solid (hint: it is always identical)? Does your calculation of NH4Cl make any sense in this context?
 
  • #3
Well, the thing that confused me is that NH4Cl is a solid, so it is not counted in the calculation for Kp. Doesn't that mean it will be [products]/0? Or do you just not divide the product by anything?

What do you mean about the solid? it says its heated, and gas forms so presumably it gets smaller. My calculations would suggest that the reaction doesn't react very far.
I'm not really clear on what Kp is. Pressure are equilibrium?
 
  • #4
Activity of solids is always 1, period. Whenever you see a solid in reaction equation, just put 1 in the reaction quotient.

How come you are asked to calculate Kp, but you don't know what it is? Please check your book or lecture notes. Seems like you are missing many basic parts of the theory, and that impedes your ability to solve the question.
 
  • #5
Ah! So I am wrong.
I assumed it was pressure at equilibrium. And after some searching I thought I had come up with the right answer.
Thanks! I'll keep looking.
 
  • #6
If, the activity of a solid is always 1.
Then would Kp =[Products]
Kp = [ 5.48*10-5] * [ 5.48*10-5]?
 

1. What is the equation for the reaction between NH4Cl and water?

The equation for the reaction between NH4Cl and water is NH4Cl + H2O ⇌ NH3 + HCl + H2O.

2. How do you calculate the Kp of the NH4Cl reaction?

The Kp of the NH4Cl reaction can be calculated by taking the partial pressure of the products (NH3 and HCl) over the partial pressure of the reactant (NH4Cl) raised to the power of their respective coefficients in the balanced equation.

3. What factors affect the Kp value of the NH4Cl reaction?

The Kp value of the NH4Cl reaction is affected by temperature, pressure, and the initial concentrations of the reactants and products. Changes in any of these factors can alter the equilibrium constant.

4. How does the Kp value relate to the spontaneity of the NH4Cl reaction?

If the Kp value is greater than 1, the reaction is spontaneous in the forward direction. If the Kp value is less than 1, the reaction is spontaneous in the reverse direction. A Kp value of 1 indicates that the reaction is at equilibrium.

5. Is the Kp value of the NH4Cl reaction affected by the presence of a catalyst?

No, the Kp value of a reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of the reaction, but does not alter the equilibrium constant or the overall amount of products and reactants at equilibrium.

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