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kow [346]
3 years ago
11

Write the expression for the equilibrium constant for the reaction represented by the equation CaCO3(s)⇌Ca2+(aq)+CO32−(aq). Is K

c > 1, < 1, or ≈ 1? Explain your answer.
Chemistry
1 answer:
Alexeev081 [22]3 years ago
5 0

<u>Answer:</u> For the given reaction, the value of K_c is greater than 1

<u>Explanation:</u>

For the given chemical equation:

CaCO3(s)\rightleftharpoons Ca^{2+}(aq.)+CO_3^{2-}(aq.)

The expression of K_c for above equation follows:

K_c=\frac{[Ca^{2+}]\times [CO_3^{2-}]}{[CaCO_3}]\\\\K_c=[Ca^{2+}]\times [CO_3^{2-}]

The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression

As, the denominator is missing and the numerator is the only part left in the expression. So, the value of K_c will be greater than 1.

Hence, for the given reaction, the value of K_c is greater than 1

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Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 16°
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PNO₂ = 0.49 atm

PN₂O₄ = 0.45 atm

Explanation:

Let's begin with the equation of ideal gas, and derivate from it an equation that  involves the density (ρ = m/V).

PV = nRT

n = m/M (m is the mass, and M the molar mass)

PV = \frac{m}{M}RT

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ρ = PxM/RT

With the density of the gas mixture, we can calculate the average of molar mass (Mavg), with the constant of the gases R = 0.082 atm.L/mol.K, and T = 16 + 273 = 289 K

2.7 = \frac{0.94xMavg}{0.082x289}

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So,

68.0687 = 46y_{NO2} + 92x(1 - y_{NO2})

68.0687 - 92 = 46y_{NO2} - 92y_{NO2}

46y_{NO2} = 23.9313

y_{NO2} = 0.52

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The partial pressure is the molar fraction multiplied by the total pressure so:

PNO₂ = 0.52x0.94 = 0.49 atm

PN₂O₄ = 0.48x0.94 = 0.45 atm

8 0
3 years ago
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