post-lab questions: 1)In this system, are the products or the reactants favored?

Question

post-lab questions: 1)In this system, are the products or the reactants favored? How does the value of Kc indicate this? Explain your answer. 2)If your lab partner used pure water instead of iron nitrate as the blank before measuring the absorbance of the solutions, are the resulting absorbance values too big, too small, or unchanged? Explain your answer. 3)If your lab partner calculated the ICE table using the concentrations of the 2.0 mM Fe+3 and SCN– solutions, neglecting the effect of dilution when the those solutions are mixed together, will the resulting K value be too big, too small, or unchanged? Explain your answer. 4)If you lab partner calculated the ICE table using millimolar units instead of molarity, would your final K value be too big, too small, or unchanged? Explain your answer

Explanation / Answer

Q1.

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Kc = [Fe(SCN)+2]/[Fe+3][SCN-]

Where Keq is constant at a given temperature, i.e. it is dependant on Temperature only

If Keq > 1, this favours products, since this relates to a higher amount of C + D

If Keq < 1, this favours reactants, since this relates to a higher amount of A + B

If Keq = 1, this is in equilibrium, therefore, none is favoured, both are in similar ratios

Note that the concentrations MUST be in equilibrium. If these are not in equilibrium, then the reaction will take place until there is equilibrium achieved.

Q2.

the solution will be LOW in absorbamce, since there is no color in solution; meaning that the light will not be "absorbed" that is, A is lower for water than Fe+3 solution

Q3

if he got

[Fe+3] = 2 mM; [SCN-] = 2m M

then, this clearly is incorrect, since it does not accounts correctly for A, which implies higher Fe+3

the K vlaue will be higher, since reactants will be accounted less

4)If you lab partner calculated the ICE table using millimolar units instead of molarity, would your final K value be too big, too small, or unchanged? Explain your answer

The effect:

Kc = [Fe(SCN)+2]/[Fe+3][SCN-]

if he used mM --> 10^-3 effect

Kc = (10^3) / (10^3)(10^3)

Kc = 1/(10^3) = 10^-3 lower

Kc value is LESS than actual vlaue

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