Consider the reaction: CCI_4(g) + 3 O_2(g) 4 CIO(g) + CO_2(g) Write the equilibr
ID: 895082 • Letter: C
Question
Consider the reaction: CCI_4(g) + 3 O_2(g) 4 CIO(g) + CO_2(g) Write the equilibrium constant expression for the reaction. Using the Thermodynamic Table provided, calculate the Gibbs Free Energy change for this reaction at 298.15K. (Delta G degree (CIO(g)) = 97.11 kJ/mol) Calculate the equilibrium constant for this reaction. Initially, 1 bar of CCI_4 and 3 bar of O_2 are added to a vessel. What is the total pressure in the system if the reaction: Goes exactly half way. Goes to completion. If the reaction vessel is compressed to a lesser volume, in which direction will this reaction shift after it has reached equilibrium?Explanation / Answer
CCl4 (g) + 3O2 (g) <-------------> 4 ClO(g) + CO2(g)
(a) K = [ClO]^4 * [CO2] / ( [CCl4] * [O2]^3 )
(b) delta Go = 4 * delta Go (ClO) + delta Go (CO2) - delta Go (CCl4)
delta Go = 4 * ( 97.11 ) - 394.6 + 58.2 kJ/mol
deltaGo = 52.04 kJ/mol
(c) deltaGo = - 2.303 RTlogKeq
52.04 x 10^3 J/mol = - 2.303 * 8.314 J/K-mol * 298.15 K * log(Keq)
log(Keq) = - 9.116
Keq = 7.66 x 10^-10
(d)
CCl4 (g) + 3O2 (g) <-------------> 4 ClO(g) + CO2(g)
1 3 0 0
1 - x 3 - 3x 4x x
Total Pressure a texactly half way = 4.5 bar
Total pressure at completion = 5 bar
Since the volume is compressed or decreased the equilibrium will shift to the direction having fewer moles of gas.
Reactant side has 4 mole of gas while product side has 5 mole of gas
So, Equilibrium shifts to the left.