Part A Given the concentrations, calculate the equilibrium constant for this rea
ID: 978098 • Letter: P
Question
Part A
Given the concentrations, calculate the equilibrium constant for this reaction:
I2(g)+Cl2(g)2ICl(g)
At equilibrium, the molar concentrations for reactants and products are found to be [I2]=0.50 M, [Cl2]=0.60 M, and [ICl]=5.0 M. What is the equilibrium constant (Kc) for this reaction?
Express your answer numerically using two significant figures.
Part B
The concentration of I2(g) is increased to 1.5 M, disrupting equilibrium. Calculate the new ratio of products to reactants with this higher concentration of iodine. Assume that the reaction has not yet regained equilibrium.
Express your answer numerically using two significant figures.
The following reaction is at equilibrium in a 2.0-L vessel:
N2(g)+3H2(g)2NH3(g) Kc=6.0×105
How do the following actions affect the equilibrium of the reaction?
Place the actions into the appropriate bins.
DECREASE VOL ,INCREASE VOL ,ADD CATALYST ,ADD NITROGEN GAS, REMOVE HYDROGEN GAS
Explanation / Answer
A.
K = [ICl]^2 / [I2][Cl2]
K = (5^2)/(0.5*0.6) = 83.33
b.
decreasing volume --> will favour the least mol formation, then NH3 is favoured (shift goes to right)
increase vol --> favours the left side (see above)
ctalyst --> no effect in euilibrium, only rates
Adding reactants N2 --> favours more production
removing reactants --> favours reactant formation