1.-The vapor pressure of water at 25 degrees celsius is 23.8 mmHg. Write Kp for

Question

1.-The vapor pressure of water at 25 degrees celsius is 23.8 mmHg. Write Kp for the vaporization of water in the unit of atm. What is Kc for the vaporization process?

2.-When 1.00 mol of I2 (g) is introduced into a 1.00L container at 1200 degrees celsius it is 5.0 percent dissociated into I (g) atoms.*Calculate Kc and Kp

3.-Hydrogen iodide decomposes according to the reaction : 2HI(g) ??H2(g) + I2(g)

At 703K a sealed 1.50 L container initially holds 0.00623 mol H2, 0.00414 mol I2, and 0.0244 mol HI at 703 K. When equilibrium is reached the concentration of H2 is 0.00467M. What are the equilibrium concentrations of HI and I2?

4.-H2 at 1.75 atm and O2 at 0.75 atm are placed in a container at 200 degrees celsius. A spark causes the reaction to form H2O(g). At equilibrium the total pressure is 1.90 atm.

Calulate Kp and Kc.

5.-The equilibrium constant for the reaction below at 986 degrees celsius is 0.63. A mixture of 1.0 mole of H2O vapor and 3.0 moles of CO is allowed to come to equilibrium at a total pressure of 2.0 atm:

CO(g)+ H2O(g) ??CO2(g)+H2(g)

How many moles H2 are present at equilibrium? and What is the partial pressure of each gas at equilibrium?

Explanation / Answer

1) H20(l) ----> H20 (g)

Kp = pH20

given pH20 = 23.8 = 23.8 /760 = 0.031315789

Kp = 0.031315789 atm

We know that   

Kp = Kc ( RT) ^ dn

here dn =1

Kc = Kp /RT

Kc = 0.031315789 / 0.0821 x 298

Kc = 1.28 x 10-3

2) I2 ----> 2 I (g)

given 5% dissociation

at equilibrium

moles of I2 = 1 - 0.05 =0.95

moles of I = 2 x 0.05 =0.1

total pressure = nRT / V = 1 x 0.0821 x 1473 / 1 = 120.933 atm

pI2 = ( 0.95 / 1.05) x 120.933 = 109.4155

pI = ( 0.1 /1.05) x 120.933 = 11.5174

Kp = ( pI)2 / pI2

Kp = ( 11.5174)2 / 109.4155

Kp = 1.21

Kc = (0.1)2/0.95

Kc = 1.05 x 10-2

3) moles of H2 = 1.5 x 0.00467 = 7.005 x 10-3

2HI -----> H2 + I2

moles of H2 fromed = 7.005 x 10-3 - 0.00623 = 7.75 x 10-4

final moles of I2 = 7.75 x 10-4 + 0.00414 = 4.915 x 10-3

conc if i2 = 4.915 x 10-3 /1.5 = 3.276 x 10-3 M

final moles of HI = 0.0244 - 2 x 7.75 x 10-4 = 0.02285

conc of HI = 0.02285 /1.5 = 0.01523333 M

4) 2H2 + 02 ----> 2H20

initial moles of H2 = 1.75 /0.0821 x 473 = 0.045

initial moles = 0.75 / 0.0821 x 473 = 0.0193

final moles = 1.90 /0.0821 x 473 = 0.048927

0.045 - 2x + 0.0193 - x + 2x = 0.048927

x = 0.015373

Kc = (H20)2 / (H2)2 (02)

Kc = 9.453 x 10-4 / 2.03 x 10-4 x 3.927 x 10-3

Kc = 1.19 x 10^3

Kp = Kc ( RT) ^-1

Kp = 30.54

5) CO(g)+ H2O(g) ??CO2(g)+H2(g)

Kc = x^2 / ( 1-x) ( 3-x)

Kc=0.63

x= 0.68

pCo = ( 2.32 / 4) x 2 = 1.16 atm

pH20 = ( 0.32/4) x 2 =0.16 atm

pC02 = ( 0.68/4) x 2 =0.34 atm

pH2 =0.34 atm

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