At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. Both the forward and the reverse reactions are occurring, but because they are occurring at the same rate, no change is observed in the appearance of the solution.

For this analysis, think about the affect that the addition or removal of reactant will have on the rate of the forward reaction relative to the rate of the reverse reaction at the instant that the reactant is added or removed. You also have to think about the addition or removal of product on the rate of the reverse reaction relative to the rate of the forward reaction.

a. When the concentration of reactant is increased for a system at equilibrium, what is the effect on the rate of the forward reaction? Does an increase in the rate of reactant have any effect on the rate of the reverse reaction? Explain your answer.

b. When the concentration of reactant is increased for a system at equilibrium, which reaction is immediately faster – the forward or the reverse reaction? Why?

c. As the reaction proceeds and the added reactant is used up, what happens to the rate of the forward reaction? At the point at equilibrium has been reached, what has happened to the rate of the forward reaction relative to the rate of the reverse reaction?

d. If reactant is removed from a system at equilibrium, what is the effect on the rate of the forward reaction? Does the rate of the reverse reaction change when reactant is removed?

e. When reactant is removed in a system at equilibrium, which reaction is faster – the forward reaction or the reverse reaction? Would that result in an accumulation of reactant or product in the flask?

f. Write a statement about why the addition of product causes the reaction to shift to reactants in terms of the relative rates of the forward and reverse reactions.

g. Write a statement from a kinetic perspective about why the removal of reactant causes the reaction to shift towards reactant.

What volume of a 0.1 M NaOH solution must be combined with 25 mL of 0.2 M HNO3 to reach a pH of 11.5?

Something that burns is said to be flammable. Do you think steel wool is flammable? is it readily flammable in air? in oxygen? what sort of differences to you observe?

In the study of biochemical processes, a common buffering agent is the weak base trishydroxymethylaminomethane, (HOCH2)3CNH2, often abbreviated as Tris. At 25 ?C,                 Tris has a pKbof 5.91. The hydrochloride of Tris is (HOCH2)3CNH3Cl, which can be abbreviated as TrisHCl. Question 1.) What volume of 10.0 M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl? ANSWER IS 6.67 Question 2.) The buffer from Part A is diluted to 1.00 L. To half of it (500. mL), you add 0.0100mol of hydrogen ions without changing the volume. What is the pH of the final solution? Answer IS 7.58 Question 3.) What additional volume of 10.0 M HCl would be needed to exhaust the remaining capacity of the buffer after the reaction described in Part B?   

Determine the pH during the titration of 25.4 mL of 0.346 M HI by 0.346 M NaOH at the following points:

(a) Before the addition of any NaOH

(b) After the addition of 12.7 mL of NaOH

(c) At the equivalence point

(d) After adding 32.0 mL of NaOH

A buffer is prepared by adding 150 mL of 1.0 M NaOH to 250 mL of 1.0 M NaH₂PO₄. How many moles of HCl must be added to this buffer solution to change the pH by 0.25 units? Assume the total volume remains unchanged at 400 mL.
For H₂PO₄^-, K_a = 6.3 × 10^-8.

Based on the data above, calculate the slope of the line of the absorbance vs relative concentration for each for the four data sets (Part III in your lab manual).

Let:

αHMR
IA equal the slope for solutions A₀-A₃ at 425 mm

αMR^-
IB equal the slope for solutions B₀-B₃ at 425 mm

αHMR
IIA equal the slope for solutions A₀-A₃ at 520 mm

αMR^-
IIB equal the slope for solutions B₀-B₃ at 520 mm

These α's are the respective Beer's Law constants for HMR and MR^- at the 2 wavelengths. Enter below the α's that you determined from your Excel graphs

Consider the dissolution of calcium hydroxide:

Ca(OH)2(s) --> Ca2+(aq) + 2OH‑(aq)

K = 4.7 x 10‑6

Write the equilibrium expression for this reaction.

2. Identify whether the following actions would favor the formation of products, reactants, or have no affect upon the dissolution of calcium hydroxide. Explain each answer.

a. Addition of CaCO3 to an equilibrium mixture of calcium hydroxide and water.

b. Addition of Na2CO3 to an equilibrium mixture of calcium hydroxide and water.

c. Addition of HCl to an equilibrium mixture of calcium hydroxide and water.

d. Addition of NaOH to an equilibrium mixture of calcium hydroxide and water

2 NO(g) + Cl₂(g) --> 2 NOCl (g)

Two proposed mechanisms for the reaction are given below. Which is the better mechanism? Explain both why the better mechanism is better and why the worse mechanism is worse.

       Mechanism 1                                                        Mechanism 2

                           NO + Cl₂ --> NOCl + Cl                                       Cl₂ --> Cl + Cl

                           NO + Cl --> NOCl NO + Cl --> NOCl

NO + Cl₂ --> NOCl + Cl

For the correct mechanism, identify:

a. the rate-limiting step

b. any intermediates

c. any catalysts

Tartaric acid is a diprotic acid (H2T) present in many fruits and is
used in cosmetics due to its powerful depigmenting effect. Therefore, it is
present in multiple facial creams whose pH is 5.2. Calculate the amount of H2T that is
you must add to prepare 125 mL of a necessary tartaric acid solution in the
manufacture of such creams. Data: Ka1 = 9.2 x 10-4; Ka2 = 4.31 x10-5. PM H2T = 150.1g / mol

The enzyme is studied in the presence of a different inhibitor (inhibitor B). In this case, two different concentrations of inhibitor are used. Data are as follows:

a) Determine the apparent Vmax at each inhibitor concentration.

b) Determine the apparent KM at each inhibitor concentration.

c) Estimate KI from these data. [Hint: Calculate the K_I at 3 mM and at 5 mM; and then take the average of the two K_{I }values.]

Sodium sorbate (HS) and sodium benzoate (HBz) are additives used together
in many foods to inhibit the growth of microorganisms. In particular, it
they are present in jams in the following amounts, sodium sorbate
(150.22 g / mol) 44 mg / kg and sodium benzoate (144.11 g / mol) 26 mg / kg. 10 g of
jam and dissolve in H2O until a total volume of 70 mL is obtained. Determines
the pH of the resulting solution assuming they are the only acid / basic species
existing in the mix. Ka sorbic acid (HS) = 1.74 x 10-5, Ka benzoic acid (HBz) =6.28 x 10-5.

You need to prepare an acetate buffer of pH 5.31 from a 0.809 M acetic acid solution and a 2.81 M KOH solution. If you have 880 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.31 ? The p K a of acetic acid is 4.76.