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denis-greek [22]
4 years ago
14

The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the stan

dard molar Gibbs energy of formation of X(g) is 5.61 kJ·mol−1 at 2000. K and −52.80 kJ·mol−1 at 3000. K. Determine the value of K (the thermodynamic equilibrium constant) at each temperature. K at 2000. K= K at 3000. K= Assuming that ΔH∘rxn is independent of temperature, determine the value of ΔH∘rxn from this data. ΔH∘rxn=
Chemistry
1 answer:
BlackZzzverrR [31]4 years ago
6 0

Answer:

The equilibrium constant at 2000 K is 0.7139

The equilibrium constant at 3000 K is 8.306

ΔH = 122.2 kJ/mol

Explanation:

Step 1: Data given

the standard molar Gibbs energy of formation of X(g) is 5.61 kJ/mol at 2000 K

the standard molar Gibbs energy of formation of X(g) is -52.80 kJ/mol at 3000 K

Step 2: The equation

1/2X2(g)⟶X(g)

Step 3: Determine K at 2000 K

ΔG = -RT ln K

⇒R = 8.314 J/mol *K

⇒T = 2000 K

⇒K is the equilibrium constant

5610 J/mol = -8.314 J/molK * 2000 * ln K

ln K = -0.337

K = e^-0.337

K = 0.7139

The equilibrium constant at 2000 K is 0.7139

Step 4: Determine K at 3000 K

ΔG = -RT ln K

⇒R = 8.314 J/mol *K

⇒T = 3000 K

⇒K is the equilibrium constant

-52800 J/mol = -8.314 J/molK * 3000 * ln K

ln K = 2.117

K = e^2.117

K = 8.306

The equilibrium constant at 3000 K is 8.306

Step 5: Determine the value of ΔH∘rxn

ln K2/K1 = -ΔH/r * (1/T2 - 1/T1)

ln 8.306 /0.713 = -ΔH/8.314 * (1/3000 - 1/2000)

2.455 = -ΔH/8.314 * (3.33*10^-4 - 0.0005)

2.455 = -ΔH/8.314 * (-1.67*10^-4)

-14700= -ΔH/8.314

-ΔH = -122200 J/mol

ΔH = 122.2 kJ/mol

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andrew-mc [135]

Answer: The molecular formula for the given organic compound X is C_6H_{8}O_7

Explanation:

We are given:

Mass of CO_2=4.13g

Mass of H_2O=1.13g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 4.13 g of carbon dioxide, =\frac{12}{44}\times 4.13=1.13g of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 1.13 g of water, \frac{2}{18}\times 1.13=0.125g of hydrogen will be contained.

Mass of oxygen in the compound = (3.00) - (1.13+ 0.125) = 1.75 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.13g}{12g/mole}=0.094moles

Moles of Hydrogen =\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.125g}{1g/mole}=0.125moles

Moles of Oxygen =\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.75g}{16g/mole}=0.109moles

Step 2: 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

For Carbon = \frac{0.094}{0.094}=1

For Hydrogen = \frac{0.125}{0.094}=1.33

For Oxygen = \frac{0.109}{0.094}=1.16

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1: 1.33: 1.16

Converting them into whole number ratios by multiplying by 6:

The ratio of C : H : O = 6: 8: 7

Hence, the empirical formula for the given compound is C_6H_8O_7

Empirical mass = 6\times 12+8\times 1+7\times 16=192g

The equation used to calculate the valency is :

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

Putting values in above equation, we get:

n=\frac{192g/mol}{192g/mol}=1

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

C_6H_8O_7\times 1=C_6H_{8}O_7

Thus molecular formula for the given organic compound X is C_6H_{8}O_7

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Nonamiya [84]

Volume of O2 : 168 L

<h3>Further explanation</h3>

Given

Reaction

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

5 moles of KCIO3

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