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Andreas93 [3]
3 years ago
14

When she introduced CO(g) and Cl2(g) into a 1.00 L evacuated container, so that the initial partial pressure of CO was 1.86 atm

and the initial partial pressure of Cl2 was 1.27 atm, she found that the equilibrium partial pressure of COCl2 was 0.823 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction.
Chemistry
1 answer:
cupoosta [38]3 years ago
5 0

Answer:

Kp is 0.00177

Explanation:

We state the equilibrium:

CO(g) + Cl₂(g)  ⇆  COCl₂(g)

Initially we have these partial pressures

1.86 atm for CO and 1.27 for chlorine.

During the reaction, x pressure has been converted. As we have 0.823 atm as final pressure in the equilibrium for COCl₂, pressure at equilibrium for CO and chlorine will be:

1.86 - x for CO and 1.27 - x for Cl₂.

And x is the pressure generated for the product, because initially we don't have anything from it. So pressure in equilibrium for the reactants will be:

1.86 - 0.823 = 1.037 atm for CO

1.27 - 0.823 = 0.447 atm for Cl₂

Let's make, expression for Kp:

Partial pressure in eq. for  COCl₂ / P. pressure in eq. for CO . P pressure in eq. for Cl₂

Kp = 0.823 / (1.037 .  0.447) → 0.00177

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The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ (n_{A}) = 1

Mole ratio of the base, NaOH (n_{B}) = 2

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH (M_{B}) = 0.505 M

Volume of the acid, H₂SO₄ (V_{A}) = 40 mL

Molarity of the acid, H₂SO₄ (M_{A}) = 0.505 M

<h3>Volume of the base, NaOH (V_{B}) =? </h3>

\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}

Cross multiply

0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4

Divide both side by 0.505

V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}

Therefore, the volume of NaOH required is 0.08 dm³

Learn more: brainly.com/question/19053582

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