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Simora [160]
1 year ago
7

A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.

Chemistry
1 answer:
kaheart [24]1 year ago
4 0

The equilibrium constant of the reaction is 282. Option D

<h3>What is equilibrium constant?</h3>

The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.

Concentration of bromine = 0.600 mol /1.000-L = 0.600 M

Concentration of iodine = 1.600 mol/1.000-L =  1.600M

In this case, we must set up the ICE table as shown;

              Br2(g) + I2(g) ↔ 2IBr(g)

I          0.6            1.6           0

C      -x                -x             +2x

E    0.6 - x         1.6 - x       1.190

If 2x = 1.190

x = 1.190/2

x = 0.595

The concentrations at equilibrium are;

[Br2] = 0.6 -  0.595 = 0.005

[I2] =   1.6 - 0.595 = 1.005

Hence;

Kc = [IBr]^2/[Br2] [I2]

Kc = ( 1.190)^2/(0.005) (1.005)

Kc = 282

Learn more about equilibrium constant:brainly.com/question/15118952

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Which of the following statements provides the BEST explanation for the many chemical similarities between two elements in a Mai
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AleksandrR [38]

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1 year ago
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5 0
3 years ago
A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a
Mandarinka [93]
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

pH = pKa + log [A]/[HA] 

 where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid

Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol

Solution:
pKa = - log ( 1.8x10^-5) = 4.74

[A] = 0.015 mol + 0.100 mol = .115 moles
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pH = 4.74 + log (.115/0.085)
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