Question

A mixture of 0.2000 mol of CO2, 0.1000 mol of H2 and 0.1600 mol of H2O is placed in a 2.000-L vessel. The following equilibrium is established at 500 K...
Co2 (g) + H2 (g) image from custom entry tool CO (g) + H2O (g)
a. Calculate the initial partial pressures of CO2, H2, and H2O.
b. At equilibrium PH2O= 3.51 atm. Calculate the equilibrium partial pressures of CO2, H2, and CO.
c. Calculate Kp for this reaction

Answer 1

Answer:

a) p(CO2) = 4.103 atm

p(H2) = 2.0515 atm

p(H2O) = 3.2824 atm

b) pCO2  = 3.8754 atm

pH2 =  1.8239 atm

pCO = 0.2276 atm

pH2O = 3.51 atm

c) Kp= 0.113

Explanation:

Step 1: Data given

Moles of CO2 = 0.2000 mol

Moles of H2 = 0.1000 mol

Moles of H2O = 0.1600 mol

Volume = 2.000 L

Temperature = 500 k

Step 2: The balanced equation

CO2 (g) + H2 (g) → CO (g) + H2O (g)

Step 3: Calculate the initial partial pressures of CO2, H2, and H2O.

pV = nRT

P = (nRT)/V

p(CO2) = (0.2000 mol * 0.08206 * 500)/2.000L

⇒ p(CO2) = 4.103 atm

p(H2) = (0.1000 mol * 0.08206 * 500)/2.000L

⇒ p(H2) = 2.0515 atm

p(H2O) = (0.1600 mol * 0.08206 * 500)/2.000L

⇒ p(H2O) = 3.2824 atm

b. At equilibrium PH2O= 3.51 atm. Calculate the equilibrium partial pressures of CO2, H2, and CO.

Step 1: Calculate the change in pH2O

The change in pH2O = 3.51 - 3.2824 = 0.2276

Step 2: the initial pressures

pCO2 = 4.103 atm

pH2 = 2.0515 atm

pCO = 0 atm

pH2O = 3.2824

Step 3: The partial pressure at the equilibrium

Since there reacts 0.2276 atm for H2O, and the mol ratio is 1:1:1:1

For each gas, there will react 0.2276 atm

pCO2 = 4.103 - 0.2276 = 3.8754 atm

pH2 = 2.0515 - 0.2276 = 1.8239 atm

pCO = 0 + 0.2276 = 0.2276 atm

pH2O = 3.51 atm

c. Calculate Kp for this reaction

Kp = (pCO * pH2O)/ (pCO2 * pH2)

Kp = (0.2276 * 3.51) /( 3.8754 *1.8239)

Kp = 0.113  

Answer 2

a). The initial partial pressures of the given compounds would be as follows:

$CO2 = 4.103 atm, H2 = 2.0515 atm, H2O = 3.2824 atm$

b). The equilibrium partial pressures would be as follows:

$CO2 = 3.8754 atm, H2 = 1.8239 atm, CO = 0.2276 atm$

c). The Kp for the given reaction would be as follows:

Kp $= 0.113$

Given that,

Mol of CO2 $= 0.2000$

Mol of H2 $= 0.1000$

Mol of H2O $= 0.1600 mol$

The Capacity of the vessel(V) $= 2 liters$

Equilibrium temperature $= 500 K$

The reaction:

$CO2 (g) + H2 (g)$ → $CO (g) + H2O (g)$

Now, the initial pressures could be calculated through the formula:

$pV = nRT$

∵ $P = (nRT)/V$

For CO2

$P = (nRT)/V$

= $(0.2000 mol$ × $0.08206$ × $500)/2.000L$

∵ Pressure for (CO2) $= 4.103 atm$

For H2

$P = (nRT)/V$

= $(0.1000 mol$  × $0.08206$ × $500)/2.000L$

∵ Pressure for H2 $= 2.0515 atm$

For H2O

$P = (nRT)/V$

= $(0.1600 mol$  × $0.08206$ × $500)/2.000L$

∵ Pressure for H2O $= 3.2824 atm$

b). To find, the equilibrium partial pressures, the mol ratio would be considered i.e. 1:1:1:1 for CO2, H2, CO, and H20.

For H2O. the reaction is $0.2276 atm$ (through differenct in pressure of H2O's pressure  $3.51 - 3.2824 = 0.2276$)

So, the equilibrium partial pressures are:

$CO2 \\= 4.103 - 0.2276 \\= 3.8754 atmH2 = 2.0515 - 0.2276 \\= 1.8239 atm$

$CO = 0 + 0.2276 \\= 0.2276 atmH2O = 3.51 atm$

c). The Kp would be calculated through

Kp $= (Pressure of CO$ × $pressure of H2O$) ÷ ($Pressure of CO2$ × $pressure of H2)$

$= (0.2276$  × $3.51)$ ÷ $( 3.8754$ × $1.8239)$

$= 0.113$

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