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Daniel [21]
3 years ago
10

Suppose the reaction system below has already reached equilibrium. Predict the effect that each of the following changes will ha

ve on the equilibrium position. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected.
UO2(s) + 4 HF(g) ↔ UF4(g) + 2 H2O(g)

1. Water vapor is removed

2. UF4(g) is added

3. The reaction is done in a glass reaction vessel. HF(g) attacks and reacts with the glass

4. If the reaction is endothermic, and you want to make the equilibrium shift to the right to maximize the products, would you heat or cool the reaction vessel?
Chemistry
1 answer:
WARRIOR [948]3 years ago
5 0

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

  • On removal of reactant from equilibrium reaction ,shifts the equilibrium in backward direction.
  • On addition of reactant from equilibrium reaction ,shifts the equilibrium in forward direction.
  • On removal of product from equilibrium reaction, shifts the equilibrium in forward direction.
  • On addition of product from equilibrium reaction, shifts the equilibrium in backward direction.

UO_2(s) + 4 HF(g)\rightleftharpoons UF_4(g) + 2 H_2O(g)

1. Water vapor is removed

On removal of water vapor from equilibrium will decrease the product which will lead to moving forward of an equilibrium that is equilibrium will shift to right direction.

2. UF_4(g) is added

On addition of UF_4(g) from equilibrium will increase the reactant which will lead to moving forward of an equilibrium that is equilibrium will shift to right direction.

3.The reaction is done in a glass reaction vessel. HF(g) attacks and reacts with the glass

Due to reaction of HF with glass will result in decrease in amount of HF present at equilibrium. So, order to counter this equilibrium will shift in backward direction or in left direction.

4. If the reaction is endothermic, and you want to make the equilibrium shift to the right to maximize the products, would you heat or cool the reaction vessel.

I endothermic reaction , heat is added to the reaction for which we can treat heat as a reactant.Now, we want to shift the equilibrium to right which means in the forward direction.

And this can be done by increasing the reactant amount, here by adding more heat to the equilibrium will shift the reaction to the right side.

So, for increasing heat we will heat the reaction vessel.

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A materials scientist has created an alloy containing aluminum, copper, and zinc, and wants to determine the percent composition
liubo4ka [24]

Answer:

0.8749 grams of hydrogen gas was formed from the reaction.

Explanation:

P = Pressure of hydrogen gad= 744 Torr = 0.98 atm

(1 atm = 760 Torr)

V = Volume of hydrogen gas= 11 L

n = number of moles of hydrogen gas= ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 27.0 °C = 300.15 K

Putting values in above equation, we get:

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.98 atm\times 11 L}{0.0821 L atm/mol K\times 300.15 K}

n = 0.4374 moles

Mass of 0.4374 moles of hydrogen gas:

0.4374 mol × 2 g/mol = 0.8749 g

0.8749 grams of hydrogen gas was formed from the reaction.

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(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct
Zarrin [17]

Explanation:

The given data is as follows.

   \Lambda^{o}_{m}(NaCl) = 1.264 \times 10^{-2}

   \Lambda^{o}_{m}(H-O=C-ONO) = 1.046 \times 10^{-2}

   \Lambda^{o}_{m}(HCl) = 4.261 \times 10^{-2}

Conductivity of monobasic acid is 5.07 \times 10^{-2} S m^{-1}

     Concentration = 0.01 mol/dm^{3}

Therefore, molar conductivity (\Lambda_{m}) of monobasic acid is calculated as follows.

                 \Lambda_{m} = \frac{conductivity}{concentration}

                                  = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}

                                 = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}

                                 = 5.07 \times 10^{-3} S m^{2} mol^{-1}

Also, \Lambda^{o}_{m} = \Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}

                            = 4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}

                            = 4.043 \times 10^{-2} S m^{2} mol^{-1}

Relation between degree of dissociation and molar conductivity is as follows.

               \alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}

                             = \frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}

                             = 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

Putting the values into the above formula we get the following.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

                        = \frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}

                        = 0.017973 \times 10^{-2}

                       = 1.7973 \times 10^{-4}

Hence, the acid dissociation constant is 1.7973 \times 10^{-4}.

Also, relation between pK_{a} and K_{a} is as follows.

                 pK_{a} = -log K_{a}

                              = -log (1.7973 \times 10^{-4})

                              = 3.7454

Therefore, value of pK_{a} is 3.7454.

                             

3 0
3 years ago
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