Asked by Troy
Suppose a 250. mL flask is filled with 1.1 mol of Cl2 and 0.60 mol of HCl. The following reaction becomes possible:
H2(g)+Cl2(g) <---> 2HCl (g)
The equilibrium constant K for this reaction is 0.931 at the temperature of the flask. Calculate the equilibrium molarity of HCl. Round your answer to two decimal places.
H2(g)+Cl2(g) <---> 2HCl (g)
The equilibrium constant K for this reaction is 0.931 at the temperature of the flask. Calculate the equilibrium molarity of HCl. Round your answer to two decimal places.
Answers
Answered by
DrBob222
(HCl) = 0.60/0.250 = approx2.5 but you need a better number than this.
(Cl2) = 1.1/0.250 = approx 4
..................H2(g)+Cl2(g) <---> 2HCl (g)
I..................0.........4.................2.5
C................x..........x..................-2x
E................x........4+x..............2.5-2x
Write the Kc expression, plug the E line into the Kc expression and solve for x, then evaluate 2.5-x. Remember to use numbers you calculate and not my estimates. Post your work if you get stuck.
(Cl2) = 1.1/0.250 = approx 4
..................H2(g)+Cl2(g) <---> 2HCl (g)
I..................0.........4.................2.5
C................x..........x..................-2x
E................x........4+x..............2.5-2x
Write the Kc expression, plug the E line into the Kc expression and solve for x, then evaluate 2.5-x. Remember to use numbers you calculate and not my estimates. Post your work if you get stuck.
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