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Simora [160]
2 years 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]2 years 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

#SPJ1

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Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

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The following mass amounts of each compound participate in the reaction:

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An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

  • P= 2 atm
  • V= 10 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 298 K

Replacing:

2 atm*10 L= n*0.082 \frac{atm*L}{mol*K} *298 K

and solving you get:

n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }

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The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

  • If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3}  }{124grams of P_{4}}

moles of PH₃=0.2742

  • If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2}  }{124grams of P_{4}}

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, <u><em>the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.</em></u>

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