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lawyer [7]
3 years ago
9

A mixture of gaseous CO and H2, called synthesis gas, is used commercially to prepare methanol (CH3OH), a compound considered an

alternative fuel to gasoline. Under equilibrium conditions at 550.3 K, [H2] = 0.07710 mol/L, [CO] = 0.02722 mol/L, and [CH3OH] = 0.0401 mol/L. What is the value of Kc for this reaction at 550.3 K?
Chemistry
1 answer:
Anit [1.1K]3 years ago
3 0

<u>Answer:</u> The value of K_c for the reaction at 550.3 K is 247.83

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

For a general chemical reaction:

aA+bB\rightarrow cC+dD

The expression for K_{c} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

The chemical equation for the production of methanol follows:

CO+2H_2\rightleftharpoons CH_3OH

The expression of K_c for above equation follows:

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

We are given:

[CH_3OH]=0.0401mol/L

[CO]=0.02722mol/L

[H_2]=0.07710mol/L

Putting values in above equation, we get:

K_c=\frac{0.0401}{0.02722\times (0.07710)^2}\\\\K_c=247.83

Hence, the value of K_c for the reaction at 550.3 K is 247.83

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Aqueous concentrated nitric acid is 69% hno3 by weight and has a density of 1.42 g/ml.
OleMash [197]

Answer: -

15.55 M

35.325 molal

Explanation: -

Let the volume of the solution be 1000 mL.

Density of nitric acid = 1.42 g/ mL

Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid

= 1000 mL x 1.42 g/ mL

= 1420 g.

Percentage of HNO₃ = 69%

Amount of HNO₃ = \frac{69} {100} x 1420 g

= 979.8 g

Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol

Number of moles of HNO₃ = \frac{979.8 g}{63 g/ mol}

= 15.55 mol

Molarity is defined as number of moles per 1000 mL

We had taken 1000 mL as volume and found it to contain 15.55 moles.

Molarity of HNO₃ = 15.55 M

Mass of water = Total mass of nitric acid solution - mass of nitric acid

= 1420 - 979.8

= 440.2 g

So we see that 440.2 g of water contains 15.55 moles of HNO₃

Molality is defined as number of moles of HNO₃ present per 1000 g of water.

Molality of HNO₃ = \frac{15.55 x 1000}{440.2}

= 35.325 molal

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Answer:

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