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Sati [7]
3 years ago
12

At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o

f HBr, H2, and Br2 are 1.0 x 10^-2 atm, 2.0 x 10^-4 atm, and 2.0 x 10^-4 atm, respecivtely, what is the equilbrium partial pressure of H2?
Chemistry
1 answer:
Damm [24]3 years ago
4 0

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

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Water is placed in a graduated cylinder and the volume is recorded as 43.5 mL. A homogeneous sample of metal pellets with a mass
Ipatiy [6.2K]

Answer:

1.8g

Explanation:

Initial volume = 43.5ml

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4 0
3 years ago
How many grams of hydrochloric acid are produced when 15.0 grams NaCl react with excess H2SO4 in the reaction
e-lub [12.9K]

Answer:

11.65 g

Explanation:

Amount of NaCl  = 15.0 grams

amount of H₂SO₄ = Excess

mass of hydrochloric acid (HCl) = ?

Solution:

To solve this problem first we will look for the reaction that NaCl react with  H₂SO₄

Reaction:

         2NaCl + H₂SO₄ --------> 2 HCl + Na₂SO₄

As the  H₂SO₄ is in excess so the amount of hydrochloric acid (HCl) depends on the amount of NaCl as it its act as limiting reactant.

Now if we look at the reaction

         2NaCl + H₂SO₄ --------> 2HCl + Na₂SO₄

           1 mol                              2 mol  

Now convert moles to mass

Then

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

So,

            2NaCl           +      H₂SO₄     -------->       2HCl        +     Na₂SO₄

      2 mol (58.5 g/mol)                                    2 mol (36.5 g/mol)

              94 g                                                           73 g

So if we look at the reaction; 94 g of NaCl gives with 73 g of hydrochloric acid (HCl) in a this reaction, then how many grams of hydrochloric acid (HCl) will produce from 15.0 g of NaCl

For this apply unity formula

           94 g of NaCl  ≅ 73 g of HCl

           15.0 g of NaCl  ≅ X g of HCl

By Doing cross multiplication

           X g of HCl = 73 g x 15.0 g / 94 g

           X g of HCl = 11.65 g

11.65 g of hydrochloric acid (HCl) will produce by 15.0 g of NaCl

7 0
3 years ago
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