l GoSmart LTE 10:44 AM * 75%- ezto.mheducation.com E connect 1HW Chaper 10, 1000
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l GoSmart LTE 10:44 AM * 75%- ezto.mheducation.com E connect 1HW Chaper 10, 1000 ponts Be sure to answer all parts. Report proble A quantity of 0.29 mole of carbon dioxide was heated to certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: C(s) + CO2(g) 2 CO g) Under these conditions, the average molar mass of the gases was 34.1 g/mol. (a) Calculate the mole fractions of CO and COz. The mole fraction of CO is The mole fraction of COz is (b)What is Kp if the total pressure is 10.7 atm? eferencesExplanation / Answer
Answer : mole fraction of CO = 0.58
mole fraction of CO2 = 0.29
Equillibrium constant Kp = 12.43 atm
Given, C(s) + CO2 (g) -----> 2CO (g)
Average molar mass of the gases = 34.1 g /mol
and no.of moles of CO2 = 0.29 moles
From, the stoichiometric equation,
1 mole of CO2 produces ----> 2 mole of CO
==> 0.29 mole of CO2 produces -----> 2* 0.29 = 0.58 mole of CO
molar mass of CO2 = 44.00 g/mol
molar mass of CO = 28.00 g/mol
total mass of CO2 = no.of moles * molar mass
= 0.29 *44 = 12.76 g of CO2
total mass of CO =no.of moles * molar mass
= 0.58 * 28 = 16.24 g of CO
Given, Average molar mass of the gases = 34.1 g /mol
mole fraction of CO = moles of CO / total moles = 0.58 / 1 = 0.58
mole fraction of CO2 = moles of CO2/ total moles = 0.29 / 1 = 0.29
Now,
Total pressure of the system = P = 10.7 atm
partial pressure of a component i = xi * P (where, xi = mole fraction of ith compound )
partial pressure of CO = pCO = 0.58 * 10.7 atm = 6.21 atm
partial pressure of CO2 = pCO2= 0.29 * 10.7 atm = 3.103 atm
C(s) do not have partial pressure as it is solid.
for a reaction
aA + bB ---> cC + dD
Kp = [pC]c [pD]d / [pA]a [pB]b
Therefore, Kp = [pCO ]2 / [pCO2] = 6.212 / 3.103 = 12.428 atm
Equillibrium constant Kp = 12.43 atm
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