A student ran the following reaction in the laboratory at 610 K: CO(g) + Cl2(g)
ID: 1066722 • Letter: A
Question
A student ran the following reaction in the laboratory at 610 K:
CO(g) + Cl2(g) COCl2(g)
When she introduced 0.183 moles of CO(g) and 0.211 moles of Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 6.72×10-2 M.
Calculate the equilibrium constant, Kc, she obtained for this reaction.
2.A student ran the following reaction in the laboratory at 546 K:
COCl2(g) CO(g) + Cl2(g)
When she introduced 0.854 moles of COCl2(g) into a 1.00 liter container, she found the equilibrium concentration of COCl2(g) to be 0.817 M.
Calculate the equilibrium constant, Kc, she obtained for this reaction.
3. Consider the following reaction:
COCl2(g) CO(g) + Cl2(g)
If 6.56×10-3 moles of COCl2, 0.377 moles of CO, and 0.372 moles of Cl2 are at equilibrium in a 16.9 L container at 772 K, the value of the equilibrium constant, Kp, is
Explanation / Answer
conc of CO = no of moles/volume in L = 0.183/1 = 0.183 M
conc of Cl2 = no of moles/volume in L = 0.211/1 = 0.211M
CO(g) + Cl2(g) ------> COCl2(g)
I 0.183 0.211 0
C -0.1438 -0.1438 0.1438
E 0.0392 0.0672 0.1438
Kc = [COCl2]/[CO][Cl2]
= 0.1438/0.0392*0.0672 = 54.6
COCl2(g) -------> CO(g) + Cl2(g)
I 0.854 0 0
C -0.037 0.037 0.037
E 0.817 0.037 0.037
KC = [CO][Cl2]/[COCL2]
= 0.037*0.037/0.817 = 0.00167
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