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1) Consider the dissolution of CaCO3 compound in aqueous medium.
a) Write down the equation of the chemical reaction that represents this dissolution.
b) Write the expression of the equilibrium constant for this reaction.
c) Explain how the addition of a certain amount of sodium carbonate to
water would affect this balance

2) The equilibrium constant for the reaction 2 SO2(g) + O2(g) 2 SO3(g) has a Kc value = 2.5x10^10 to 500 K. Find the kc value for each of the following reactions at the same temperature
(a) SO2(g) +1/2 O2 SO3(g).
(b) SO3(g) SO2(g)+1/2 O2(g)
(c) 3SO3(g)+ 3/2 O2(g) 3SO3(g)

3) A reaction mixture consisting of 0.400 mol H 2 and 1.60 mol I 2 was prepared in a 3.00 L flask and heated. In balance, 60.0% of the hydrogen gas reacted. What is the equilibrium constant for the reaction H 2(g) + I 2(g) 2 HI(g) at this temperature?

Respuesta :

MathPhys

Explanation:

1a) CaCO₃(s) → Ca²⁺(aq) + CO₃²⁻(aq)

1b) Remember, solids are not included in the equilibrium equation.

K = [Ca²⁺] [CO₃²⁻]

1c) Adding CO₃²⁻ ions will shift the reaction to the left, producing CaCO₃(s) until equilibrium is restored.

2) 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Kc = 2.5×10¹⁰ = [SO₃]² / ([SO₂]² [O₂])

2a) SO₂(g) + ½ O₂(g) → SO₃(g)

Kc = [SO₃] / ([SO₂] [O₂]^½)

Kc² = [SO₃]² / ([SO₂]² [O₂])

Kc² = 2.5×10¹⁰

Kc ≈ 1.58×10⁵

2b) SO₃(g) → SO₂(g) + ½ O₂(g)

Kc = [SO₂] [O₂]^½ / [SO₃]

Kc = 1 / (1.58×10⁵)

Kc ≈ 6.33×10⁻⁶

2c) 3 SO₂(g) + ³/₂ O₂(g) → 3 SO₃(g)

Kc = [SO₃]³ / ([SO₂]³ [O₂]^³/₂)

Kc = ([SO₃] / ([SO₂] [O₂]^½))³

Kc = (1.58×10⁵)³

Kc ≈ 3.95×10¹⁵

3) H₂(g) + I₂(g) → 2 HI(g)

K = [HI]² / ([H₂] [I₂])

Make an ICE table.

[tex]\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\H_{2}&0.400&-0.240&0.160\\I_{2}&1.60&-0.240&1.360\\HI&0&+0.480&0.480\end{array}\right][/tex]

K = (0.480)² / (0.160 × 1.360)

K = 1.06

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