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Schach [20]
3 years ago
5

One of the many reactions which occur when iron ore is smelted in a blast furnace is given below. For this reaction, the equilib

rium constant in terms of partial pressures, Kp, is 0.900 at 873 K and 0.396 at 1273 K. Assume the reaction takes place in a vessel containing only product and reactant molecules and that AH and AS do not change with temperature. FeO(s) + CO(g) = Fe(s) + CO2(g) a) Calculate AH, AG and AS for the reaction at 873 K b) What is the mole fraction of CO2(g) at 873 K? Include only the gas-phase species in this mole fraction.
Chemistry
2 answers:
Mumz [18]3 years ago
5 0

Answer:

you looking beautiful

nata0808 [166]3 years ago
4 0

Answer:

Explanation: AH, AG and AS for the reaction at 873 K

delta G =-RT ln Kp

G= -8.314 x 873 ln 0.9

= 764.7J/mol

AG = AH-TAS

AH = 764.7 - 8314(0.9)

Kp =pCO/pCo2

Substitute the values at different temp. Solve simultaneously

Since mole fraction =kp/pCo

Mf =1.5

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Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is comb
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<u>Answer:</u> The molecular formula for the menthol is C_{10}H_{20}O

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=0.2829g

Mass of H_2O=0.1159g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.2829 g of carbon dioxide, \frac{12}{44}\times 0.2829=0.077g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1159 g of water, \frac{2}{18}\times 0.1159=0.0129g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00066 moles.

For Carbon = \frac{0.0064}{0.00066}=9.69\approx 10

For Hydrogen  = \frac{0.0129}{0.00064}=19.54\approx 20

For Oxygen  = \frac{0.00066}{0.00066}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is C_{10}H_{20}O_1=C_{10}H_{20}O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

n=\frac{156.27g/mol}{156g/mol}=1

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O

Thus, the molecular formula for the menthol is C_{10}H_{20}O

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