Question

A 2.50 mole sample of NOCl was placed in a 1.50-L container at 400oC. When equilibrium has established, it was found that 28% NOCl has dissociated according to the reaction:
2NOCl(g) ⇄ 2NO(g) + Cl2(g)
Calculate the equilibrium constant Kc

Answer 1

In general

aA + bB ⇄ cC + dD

where lower case letters represent the number of atoms or molecules used in the reaction and upper case represent the identity of the atom or molecule itself

to find the equilibrium we use the equation

\[k_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]

Note the square brackets denote Molarity so when you see an element or a molecular formula in square brackets, it means the molarity of the compound

for example,

if you know the molarity of boron you reacted is 0.1M so,
[B] = 0.1M

So first thing you need to do is find the is determine the Molarity of all reactants and products, then you need to sub them into the formula. Make sure to sub in the number of atoms used in the reaction you should have a formula like this.

\[k_c = \frac{[NO]^2[Cl_2]^{1}}{[NOCl]^2}\]

This will give you the unitless equilibrium constant once you sub in the molarities of all compounds present in the reaction in their specific spot

Hope this helps and if you have any questions dont hesitate to ask

Answer 2

so the mole of NoCl that we have to use is 2.5 mole or 28% x 2.5?

Answer 3

You are kind of on the right track let me explain it further,

To find molarity you use the formula

\[Molarity = \frac{Moles}{Litres}\]

You know that, "A 2.50 mole sample of NOCl was placed in a 1.50-L"
from this you should be able to find the Molarity of NOCl

You aren't given the individual molarity, or moles of both the product and reactant, but you are given the amount of moles of total reactant used in the reaction by the remark, "28% NOCl has dissociated".

To determine the amount of moles of reactant used to produced product you have to convert 28% to a decimal as it is in units of (percent which is mostly just used because people like seeing whole numbers rather than decimals) so that you can use it to figure out the moles of product produced.

so,

28%/100 = 0.28
0.28*2.50mol = Moles of Product produced

So,

0.28*2.50mol = 0.7mol of reactant used

Now you need to figure out the moles of NO and Cl2 produced from 0.7mol

To do this you simply need to look at the formula, also you need to remember that moles can change in a reaction because moles are based on the number of particles of 12g of carbon 12.

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

so we have 0.7mol of reactant that went to form product so to find out the amount of Cl2 produced we just apply the formula,

\[(Moles of NOCl)*\frac{CoefficientOfCl_2}{Coefficient Of NOCl}\]

so,

\[(0.7mo)\frac{1}{2} = 0.35Moles of Cl_2 Produced\]

for NO you do the same thing,

\[0.7mol \frac{2}{2} = 0.7mol\]

Now you can convert both these numbers to molarity remembering that you have a total of 1.50L of solution and find your solution.

BONUS!!!
------------------------------------------------------------------------
Just for fun though I'm going to show that the mass is conserved in this reaction which confirms my results, as mass should always be conserved in a reaction

remember that,

\[Mole = \frac{Mass}{Molecular Mass}\]

So the Moles of NOCl reacted was 28% or 0.7mol and the molecular mass of NOCl (we find this using the periodic table) is,
14.0067g/mol + 15.9994g/mol + 35.453g/mol
= 65.46g/mol

so,
0.7mol*65.45g/mol = 45.8g

so we reacted 45.8g of NOCl
and we got,

0.7mol of NO
and
0.35mol of Cl2

Lets use the formula above to solve for the grams of both of these,

0.7mol*30.0061g/mol = 21.0g
0.35mol*70.906g/mol = 24.8g

so,

21.0g + 24.8g = 45.8g

Thus,
Mass of Reactant = Mass of Products
45.8g = 45.8g

Therefore, as expected the Mass was conserved and the problem was conducted correctly



I Hope this was helpful if you have any questions please do not hesitate to ask

Answer 4

I got 0.23 for Kc. Can you check for me?

Answer 5

the answer should be 0.03. Can you help me, please?

Answer 6

HINT:
Did you use the proper Molarity for NOCl?

Think about it an equilibrium involves the extent a reaction will go forward converting the TOTAL amount of reactants to products, until both the products and reactants in the reaction are in optimal concentration. Kc is just a measure of this.

Important:

0.03 is incorrect, the answer is actually less than that, the people who gave you that answer made a pretty big mistake when calculating the answer.

The error they made involves setting up the coefficients as exponents in the equation


I'm sorry for not just giving you the answer but I feel like if you ever see this on a test it is good to have an understanding of what is actually going on so that you can complete it in time and without error.

Answer 7

Oh btw, if you post your final answer I will inform you if you are correct :)

Answer 8

My answer is 0.0259. Is it correct?

Answer 9

Sorry that is incorrect, I'm not sure how you got that? The answer you are looking for is actually lower than that.

When you are solving this write all your variable down in a list

0.7mol of NO
0.35mol of Cl2
2.5mol of NOCl
Volume of 1.5L (we assume that the volume remains the same at equilibrium)

Given this information you should be able to convert each to molarity, which you can sub into the equilibrium equation that you set up using the chemical equation:

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

REMEMBER the coefficients in front of the molecular formulas of these chemicals become exponents in the equilibrium equation.

If you cannot get this correct I will show you a full solution ASAP :)
I have confidence you can get it though :)

Answer 10

Ohh.. I see. So we use 2.5 mole for NoCl. I thought we should use the mole after 28% has dissociated. So, the answer is 0.0176. Am I right?

Answer 11

I'm sorry I made a mistake

The Moles of Reactant at equilibrium would be

2.5mol - 0.7mol = 1.8mol
Thus you have 1.8 mol of NOCl

0.3 is the correct answer

sorry for that :)

Answer 12

Moles Of NOCl - Moles of NOCl Used for products = Moles of NOCl at equilibrium

Here is the solution
https://www.wolframalpha.com/input/?i=%28%280.35%2F1.5%29%280.7%2F1.5%29^2%29%2F%281.8%2F1.5%29^2

Hope I was helpful sorry about the mistake

Answer 13

Its ok. Thanks for all the help :)

Answer 14

No problem if you ever have a chemistry problem feel free to ask.