ANSWER:
Given,
number of moles COCl2 = 1.68 moles
volume of container = 1.0 L
[COCl2]initial = (1.68 mol)/(1.0 L) = 1.68 M
[Cl2]equilibrium = 3.61 x 10-2 M
reaction is :
COCl2 (g) CO (g) + Cl2 (g)
initial 1.68 M 0.0 M 0.0 M
at equili (1.68 - x) M x M x M
So, x = 3.61 x 10-2 M
[COCl2]euilibrium = (1.68 - 3.61 x 10-2 ) M = 1.644 M
[CO]equilibrium = 3.61 x 10-2 M
[Cl2]equilibrium = 3.61 x 10-2 M
Now,
Kc = 7.93 x 10-4
A student ran the following reaction in the laboratory at 529 K: CoCl2(g) P CO(g) +...
A student ran the following reaction in the laboratory at 557 K: CO(g) + Cl2(g) = COCl2(g) When she introduced 1.06 moles of CO(g) and 1.09 moles of Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of COCl2(g) to be 1.02 M. Calculate the equilibrium constant. Ko she obtained for this reaction. Ke=
1. A student ran the following reaction in the laboratory at 632 K: 2HI(g) ->H2(g) + I2(g) When she introduced 0.362 moles of HI(g) into a 1.00 liter container, she found the equilibrium concentration of I2(g) to be 3.55×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc = 2. A student ran the following reaction in the laboratory at 616 K: CO(g) + Cl2(g) -> COCl2(g) When she introduced 0.131 moles of CO(g) and 0.161 moles...
A student ran the following reaction in the laboratory at 541 K: COC12(E) CO(g) + Cl2(g) When she introduced 1.13 moles of COC12(e) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 3.83x10-2 M. Calculate the equilibrium constant, K, she obtained for this reaction.
A student ran the following reaction in the laboratory at 425 K: PCl5(g) --> PCl3(g) + Cl2(g) When she introduced 4.59 moles of PCl5(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 3.94×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.
A student ran the following reaction in the laboratory at 254 K: 2NO(g) + Br2(g) ⇌ 2NOBr(g) When she introduced 0.185 moles of NO(g) and 0.130 moles of Br2(g) into a 1.00 liter container, she found the equilibrium concentration of NOBr(g) to be 0.152 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =
A student ran the following reaction in the laboratory at 225 K: 2NOBr(g) 2 2NO(g) + Brz(g) When she introduced 0.198 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of Br2(g) to be 1.89x10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =
A student ran the following reaction in the laboratory at 651 K: 2NH3(g)N2(g) + 3H2(g) When she introduced 7.88x102 moles of NH (g) into a 1.00 liter container, she found the equilibrium concentration of NH (g) to be 6.75x103 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc
A student ran the following reaction in the laboratory at 293 K: 2CH2Cl2(g) CH4(g) + CCl4(g) When she introduced 6.91×10-2 moles of CH2Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of CCl4(g) to be 3.19×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc=?
A student ran the following reaction in the laboratory at 283 K: 2CH2Cl2(g) ->CH4(g) + CCl4(g) When she introduced 7.70×10-2 moles of CH2Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of CCl4(g) to be 3.59×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =
A student ran the following reaction in the laboratory at 324 K: 2NO(g) + Br2(g) 2NOBr(g) When she introduced 0.137 moles of NO(g) and 0.119 moles of Br2(g) into a 1.00 liter container, she found the equilibrium concentration of Br2(g) to be 6.92×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.