2 Cl2 (g) + 2 H2O (g)
O2 (g) + 4 HCl (g)
Given: ∆G° = 11.2 KJ/mol
∆H° = 114.4 KJ/mol
T = 800 K
(a) To calculate equilibrium constant
∆G° = -R T ln K
11200 J/mol = -( 8.314 J k-1 mol-1) x 800 x 2.303 log K
log K = - 11200/(8.314 x 800 x 2.303)
= - 0.7311
K = 10-0.7311
K = 0.1857
b) Given: Cl2 = 3.6 bar, H2O = 1.4bar, O2 = 0.20 bar, HCl= 0.50 bar
∆G = ∆G° + RT lnQ ------(1)
Reaction quotient (Q) = product/ reactant
According to above balanced reaction
Q = (0.20) x (0.50)4 /(3.6)2 x (1.4)2
Q = 4.9 x 10-4
ln Q = ln 4.9 x10-4
= -7.6211
Putting values in (1)
∆G = (11200 J/mol) + (8.314 x 800 x -7.6211)
= - 39.48 KJ/mol
The reaction: 2 Cl2(g) + 2 H2O(g) ➝ O2(g) + 4 HCl(g) has ΔG° = +11.2...
A mixture of 0.008603 mol of Cl2, 0.05744 mol of
H2O, 0.05353 mol of HCl, and 0.06453 mol of
O2 is placed in a 1.0-L steel pressure vessel at 1529 K.
The following equilibrium is established:
2 Cl2(g) + 2 H2O(g) 4 HCl(g) + 1 O2(g)
At equilibrium 0.006795 mol of Cl2 is found in the
reaction mixture.
(a) Calculate the equilibrium partial pressures of Cl2,
H2O, HCl, and O2.
Peq(Cl2) = .
Peq(H2O) = .
Peq(HCl) = .
Peq(O2) = .
(b) Calculate...
A mixture of 0.1468 mol of Cl2, 0.04712 mol of H2O, 0.02609 mol of HCl, and 0.1672 mol of O2 is placed in a 1.0-L steel pressure vessel at 579 K. The following equilibrium is established: 2 Cl2(g) + 2 H2O(g) 4 HCl(g) + 1 O2(g) At equilibrium 0.006984 mol of HCl is found in the reaction mixture. (a) Calculate the equilibrium partial pressures of Cl2, H2O, HCl, and O2. Peq(Cl2) = ? Peq(H2O) = ? Peq(HCl) = ? Peq(O2)...
For the reaction C2H4(g) + Cl2(g) C2H4C12(1), determine AH°, given that 2 Cl2(g)2 H2O(I) 4 HCl(g)O2(g) -202.4 kJ mol1 1 2 HCI(g) + C2H4(g) +O28) 1 -318.7 kJ mol C2H4C12(1) H2O(1) AH°
A reaction has an equilibrium constant of Kp=0.061 at 27 ∘C. Part A Find ΔG∘rxn for the reaction at this temperature. Find for the reaction at this temperature. 6.98 kJ 0.839 kJ -6.98 kJ 0.628 kJ Part B Above what temperature does the following reaction become nonspontaneous? 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g) ΔH = -1036 kJ; ΔS = -153.2 J/K 298 K 158.7 K 6.762 × 103 K This reaction is nonspontaneous at all...
For the reaction C2H4(8) + Cl2(g) — C2H4C12(1), determine A,Hº, given that 4 HCl(g) + O2(g) — 2012(g) + 2 H2O(1) AH° = –202.4 kJ mol-1 2 HCl(g) + C2H4(8) + + O2(g) — C2H4Cl2(1) + H2O(1) ArHº = -318.7 kJ mol-1
For the reaction H2(g) +
Cl2(g) 2 HCl(g), the
equilibrium constant K at 800oC is 4.35 x
104.
Hydrogen and chlorine, each at a partial pressure of 0.700 bar,
are placed in a vessel at 800oC and allowed to
equilibrate. Find the final partial pressures of all three gases in
this reaction.
p(H2) = bar
p(Cl2) =
. bar
p(HCl) =
bar
Consider the reaction shown. 4 HCl(g) + 02(g) 2 CL2(g)
+ 2 H2O(g) Calculate the number of grams of Cl2 formed when 0.375
mol HCl reacts with an excess of o2
Consider the reaction shown. 4 HCl(g) + O2(g) + 2Cl2(g) + 2 H,0(g) Calculate the number of grams of CI, formed when 0.375 mol HCl reacts with an excess of o, bo
For the reaction H2(g) + Cl2(g) 2 HCl(g), the equilibrium constant K at 800oC is 4.35 x 10^4. Hydrogen at a partial pressure of 0.700 bar and chlorine at a partial pressure of 0.500 bar are placed in a vessel at 800oC and allowed to equilibrate. Find the final partial pressures of all three gases in this reaction in bar
Calculate the equilibrium concentrations of H2O, Cl, HCl, and O2 at 298 K if the initial concentrations are (H2O) = 0.070 M and (Cl2] = 0.120 M. The equilibrium constant Kc for the reaction H2O(g) + Cl2(g) + 2HCl(g) + O2(g) is 8.96 x 10 -9 at 298 K
Consider the following reaction 4HCl(g) + O2(g) 2H2O(g) + 2Cl2(g) HCl(g) H2O (g) Hof (kJ/mol) - 92.3 - 241.8 Gof (kJ/mol) - 75.3 - 228.6 a) (12 pts) Predict which direction, forward or reverse, is accompanied by increasing disorder (Show your calculations). b) (3 pts) Can this reaction be reversed by changing the temperature? Explain. c) (3 pts) Calculate Kp for the above reaction at 25oC? d) (5 pts) Calculate the free energy, G, for the above reaction at 200C...