In a sealed 10.5-L vessel at 184C, equilibrium is establishe
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In a sealed 10.5-L vessel at 184C, equilibrium is established...
Ammonium carbamate
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Solution Summary
The solution provides detailed explanations and instructions for the problem. A chemical equilibrium in a sealed vessel is examined. THe molecular weight of the corresponding compound is determined.
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Please see the attached file for complete solution. Below is a rough copy of the content in the file.
Molecular weight of:
NO2 = 14 + 16*2 = 46 (g/mol)
NO = 14 + 16 = 30 (g/mol)
O2 = 32 (g/mol)
You will need the number of mol of corresponding substance:
nNO2 = 1.353/46 = 0.029 (mol)
nNO = 0.096/30 = 0.0032 (mol)
nO2 = 0.0512/32 = 0.0016 (mol)
To determine Kp you will need the pressure of each individual gas:
Using ideal gas law:
PV = nRT
For NO2:
For NO: pNO = 0.0032*361.86 = 1.16 (kPa)
For O2: pO2 = 0.0016*361.86 = 0.58 (kPa)
Thus
Reference:
http://en.wikipedia.org/wiki/Universal_gas_constant
Write down the balanced equation:
NH2COONH4 (s) = 2NH3 + CO2
According to stoichiometry: pNH3 = ...
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