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3 years ago

Consider the following chemical equilibrium:

Chemistry

1 answer:

Keith_Richards [23]3 years ago

7 0

Answer:

Kp = Kc (RT) ^(-2)

Explanation:

For the reaction;

N2 (g) + 3H2 ⇄ 2NH3(g)

We can write;

Kc = [NH3]^2/[N2] [H2]^3

But

Kp = pNH3^2/pN2 . PH2^3

To convert from Kc to Kp

Kp = Kc (RT) ^Δn

where Δn is the change in number of moles going from reactants

to products.

For this reaction;

Δn = 2- (3+1) = -2

Kp = Kc (RT) ^(-2)

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A measure of kinetic energy of particle motion within a substance is temperature.

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3 years ago

Hydrogen sulfide,H2S, is a very toxic gas with a smell of rotten eggs. using the following: H2S+3/2 O2=SO2+H2O H2+1/2O2=H2O S+O2

Serga [27]

Answer:

ΔH = -20kJ

Explanation:

The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:

H₂(g) + S(g) → H₂S(g)

Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:

(1) H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ

(2) H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

(3) S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

The sum of -(1) + (2) + (3) gives:

-(1) SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

(2) H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

(3) S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

-(1) + (2) + (3): H₂(g) + S(g) → H₂S(g)

ΔH = +519kJ - 242kJ - 297kJ = -20 kJ

I hope it helps!

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3 years ago

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (M) is calculated from the following equation:

M = n ÷ V

where n is the number of moles of the solute in the solution, and V is the volume of the solution.

Accordingly, the number of moles of the solute is given by

n = M x V

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

n (1) = M (1) x V (1)

n (2) = M (2) x V (2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

n (1) = n (2)

and

M (1) x V (1) = M (2) x V (2)

For the stock solution, M (1) = 1.244 M, and V (1) is the volume needed. For the dilute solution, M (2) = 0,1281 M, and V (2) = 250.0 mL.

The volume of the stock solution needed, V (1), is calculated as follows:

V (1) = M (2) x V (2) ÷ M (1)

V (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

V (1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

7 0

3 years ago

A solution containing CaCl2 is mixed with a solution of Li2C2O4 to form a solution that is 3.5 × 10-4 M in calcium ion and 2.33

Burka [1]

Answer:

B. A precipitate will form since Q > Ksp for calcium oxalate

Explanation:

Ksp of CaC₂O₄ is:

CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻

Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:

Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹

In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.

Replacing in Ksp formula:

[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.

If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.

Thus, right answer is:

B. A precipitate will form since Q > Ksp for calcium oxalate

4 0

3 years ago