Question

The equilibrium constant, Kp, for the following reaction is 1.57 at 600 K: CO(g) + Cl2(g) COCl2(g) Calculate the equilibrium partial pressures of all species when CO and Cl2, each at an intitial partial pressure of 1.70 atm, are introduced into an evacuated vessel at 600 K.

PCO = atm

PCl2 = atm

PCOCl2 = atm

B. The equilibrium constant, K_c, for the following reaction is 2.90×10^-2 at 1150 K.

2SO₃(g) 2SO₂(g) + O₂(g)
Calculate K_c at this temperature for the following reaction:
SO₂(g) + 1/2O₂(g) SO₃(g)

K_c =

C. A student ran the following reaction in the laboratory at 664 K:
2NH₃(g) N₂(g) + 3H₂(g)
When she introduced 7.30×10^-2 moles of NH₃(g) into a 1.00 liter container, she found the equilibrium concentration of H₂(g) to be 0.102 M.
Calculate the equilibrium constant, K_c, she obtained for this reaction.
K_c =

D. For the gas phase decomposition of chloroethane,
CH₃CH₂ClC₂H₄ + HCl

the rate constant in s^-1 has been determined at several temperatures. When ln k is plotted against the reciprocal of the Kelvin temperature, the resulting linear plot has a slope of -3.06×10⁴ K and a y-intercept of 33.6.
The value of the rate constant for the gas phase decomposition of chloroethane at 713 K is ____ s^-1. (answer in one significant fig.)

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Answer 1

0 Co (g) + Cl (g) - coco (9) Kp = 1,57 Т= 6oo K CO(g) + I (g). 1.70 1.7o cocly (9) tro (Partial Pressuses) Р tzte letop 1.torp People Kp : (Pco ) (Pelo) 1.57 Р (1.7-P(1.70P) Р 1.57 p2-3.4 P + 2.89 1.57 P²-5.34P + 4.54 p= On solving this quadratic equatish; 1.68 atri Equilibrium partial pressures are: 'P co 1.70-P 170 - 1668 20.02 atm 0,02 atom 1,70 - 1.68 = clo P a 1.70p Pools 1.68 atu CS Scanned with CamScanner