CHM 202-001 HW Chapter 14 instructions help Question 10 (of 15) Save & Exit 11 S
ID: 554015 • Letter: C
Question
CHM 202-001 HW Chapter 14 instructions help Question 10 (of 15) Save & Exit 11 Submit 10 1000 point 5 attempts leth Check my work 2 Be sure to answer all parts. Report problem Hint A quantity of 0.29 mole of carbon dioxide was heated to a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: Solution Guided Solution C(s) + CO2(g) 2 CO(g) Under these conditions, the average molar mass of the gases was 36.0 g/mol. (a) Calculate the mole fractions of CO and CO2. The mole fraction of CO is The mole fraction of CO2 is (b)What is Kp if the total pressure is 10.7 atm?Explanation / Answer
The reaction is
C(s) + CO2(g) -----> 2CO(g)
Molar mass of CO = 12 + 16 = 28 gm/mol
Molar mass of CO2 = 12 + 2 * 16 = 44 gm/mol
Let us assume x moles of 0.29 mole of CO2 decompose, then 2x moles of CO will be formed
Total gaseous mole = 0.29-x + 2x = 0.29 + x
Hence the equation will be
44 * (0.29-x/0.29+x) + 28 * (2x/0.29-x) = 36
Since 36 is the mid-point of the (44+28), hence we get
2x = 0.29 - x
x = 0.29/3 = 0.096667
a) mole fraction of CO = 2 * x/(0.29+x) = 1/2 = 0.5
b) mole fraction of CO2 = (0.29-x)/(0.29+x) = 1/2 = 0.5
c)
The information is insufficient in my knowledge to decipher the value of Kp