must answer all of the questions and show all of your work clearly. 1) At high t
ID: 1027151 • Letter: M
Question
must answer all of the questions and show all of your work clearly. 1) At high temperatures, a dynamic equilibrium exists between carbon monoxide, carbon dioxide, and solid carbon. co2 (g) + C (s)2CO (g),aHP= 172.5 kJ At 850 °C, Kc is 0.153. A) Calculate the value of Kp B) In a 2.0 L container, 0.500 mol CO2 is mixed with 1.00 mol C. After equilibrium is reached, 0.40 mol of CO2 remains in the container. What is the molar concentration of CO in the container? C) In a separate 1.0 L container, 0.500 mol of CO, is mixed with 1.00 mol C and 0.250 mol CO. Once mixed, will more reactants be formed to reach equilibrium, will more products be formed to reach equilibrium, or is the system already at equilibrium? What is the equilibrium constant with regards to concentration at 850 °C for the reaction CO (g-1% CO2 (g) + ½ C (s)? D) KC0.153 8) 0.25 MExplanation / Answer
1.
Kp= KC*(RT)deltan, deltan= change in no of moles of gases during the reaction = 2( moles of CO)- 1( moles of CO2)= 1, R= 0.0821 L.atm/mole.K and T= 850 deg.c= 850+273=1123K
KP= (PCO)2/ PCO2= 0.153
Given KC= 0.153, Kp = 0.153*(0.0821* 1123) = 14.10
2. Concentration= moles/Liters, concentration of CO2= 0.5/2=0.25M
KC= [CO]2/[CO2], let x= drop in concentration of CO2 to reach equilibrium.
Moles of CO at equilibrium= 0.4, molar concentration of CO at equilibrium= 0.4/2=0.2M
3. Concentrations : CO2=0.5/1=0.5M and CO=0.25/1=0.25M( the activity of C is unity and hence does not come in the equation of equilibrium constant).
Q= [CO]2/ [CO2] = (0.25)2/0.5= 0.125<Kc, the system is at equilibrium if Q=K and if Q<K, the reaction proceeds in the forward direction so as to make more CO. More products are formed
For the reaction, CO(g) ----->0.5CO2+0.5C, KC’= [CO2]0.5/ [CO]=Sqrt(KC)= sqrt(0.153)=0.3911
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