The reaction CO2( g )+C( s )2CO( g ) has K p=5.78 at 1200 K. Part A Calculate th

Question

The reaction CO2(g)+C(s)2CO(g) has Kp=5.78 at 1200 K.

Part A

Calculate the total pressure at equilibrium when 4.73 g of CO2 is introduced into a 10.0-L container and heated to 1200 K in the presence of 3.27 g of graphite.

Express your answer to three significant figures and include the appropriate units.

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Part B

Repeat the calculation of part A in the presence of 0.50 g of graphite.

Express your answer to three significant figures and include the appropriate units.

The reaction CO2(g)+C(s)2CO(g) has Kp=5.78 at 1200 K.

Part A

Calculate the total pressure at equilibrium when 4.73 g of CO2 is introduced into a 10.0-L container and heated to 1200 K in the presence of 3.27 g of graphite.

Express your answer to three significant figures and include the appropriate units.

Ptotal=

SubmitRequest Answer

Part B

Repeat the calculation of part A in the presence of 0.50 g of graphite.

Express your answer to three significant figures and include the appropriate units.

Ptotal=

Explanation / Answer

mass of CO2= 4.73 gm, moles of CO2(n)= mass/molar mass = 4.73/44=0.1075

V= 10L and T= 1200K, P= nRT/V= 0.1075*0.0821*1200/10=1.06 atm

mass of C= 3.27gm, moles of C= mass/molar mass= 3.27/12=0.2725

theoretical molar ratio =1:1 ( as per the reaction )

molar ratio of C:CO2=0.2725:0.1075 =0.2725/0.1075: 0.1075/0.1075= 2.53:1

CO2 is limiting reactant. Let P= drop in pressure of CO2 to reach equilibrium

at Equilibrium,PCO=1.06-P and CO=2P

Kp= (2P)2/(1.06-P)=5.78, 4P2/(1.06-P)= 5.78, P2/(1.06-P)= 1.445

when sovled using excel, P=0.71 atm, so at equilibrium, PCO= partial pressure of CO=2*0.71= 1.42 and that of CO2= 1.06-0.71=0.35, so total pressure = 1.42+0.35=1.77 atm

when 0.5 gm of graphit e is used, that is the limiting reactant since moles of graphire=0.5/12=0.042

moles of CO2= 0.1075

assume that all the C is consumed, when all the C is consumed, moles of CO2 at equilibrium = 0.1075-0.042=0.0655

moles of CO= 0.042*2=0.084

Q at the begining of the reaction = (PCO)2/ PCO2= 0, since there is no CO at the begining.

from gas law, moles of CO2= 0.0655*0.0821*1200/10 atm =0.645 atm and that of CO= 0.084*0.0821*1200/10=0.83 atm

Q= (PCO)2/PCO2= 0.83*0.83/0.645= 1.06<KP, but due to limting reactant C, the reaction does not reach equilibrium.

but pressure at this point after all of the C is consumed= 0.645+0.83=1.475 atm

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