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Question # 7.23, Calculate Kc for this reaction at the above temperature.

Question # 7.23, Calculate Kc for this reaction at the above temperature.

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Grade:11

1 Answers

Arun
25750 Points
5 years ago
At equilibrium: C   +   CO2  ↔    2CO          9.45g         90.55g Number of moles of CO2   = 9.45/44 = 0.21 mole Number of moles of CO = 90.55/28 = 3.23 mole   Kp=( nCO)2 /nCO2 X [p/ ∑n]1 Here p= pressure= 1 atm ∑n = Number of moles of CO + number of moles of CO2= 3.23 + 0.21 = 3.44 moles = 3.23 X 3.23/0.21 X 1/3.44 = 14.44atm Now Kc = KP/(RT)∆nFor a given reaction ∆n = 2-1 =1=14.44\0.0821X1127]1= 0.156 mol litre-1

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