Question

pre-lab assignment
answer these questions before you come to lab. show your work fully and clearly.

1. a sample was prepared by mixing 15.0 ml of $6.00 \times 10^{-5}$ m crystal violet (cv) with 5.00 ml of 0.250 m naoh. calculate the resulting concentrations of cv and of $oh^-$.

1. suppose that trial i in this lab was run using 0.10 m naoh, producing a linear kinetics graph with best-fit equation $y = -0.003008 + 0.785$. if trial ii was run using 0.30 m naoh, what would the slope of the same graph be if the reaction were (a) zero order, (b) first order, or (c) second order?

1. the rate of the reaction $y + z \

ightarrow yz$ is studied much like the reaction in this lab, except that both y and z absorb visible light; thus, their concentrations can both be measured directly in a spectrophotometer. in trial 1, z is held constant and y is measured over time, producing the top three graphs. in trial 2, y is held constant and z is measured over time, producing the bottom three graphs. based on these graphs, write the rate law for this reaction.

Explanation:

Step1: Calculate total solution volume

$V_{\text{total}} = 15.0\ \text{mL} + 5.00\ \text{mL} = 20.0\ \text{mL}$

Step2: Find diluted CV concentration

Use $M_1V_1 = M_2V_2$:
$M_{CV,2} = \frac{M_{CV,1}V_{CV,1}}{V_{\text{total}}} = \frac{6.00 \times 10^{-5}\ \text{M} \times 15.0\ \text{mL}}{20.0\ \text{mL}}$
$M_{CV,2} = 4.50 \times 10^{-5}\ \text{M}$

Step3: Find diluted $\text{OH}^-$ concentration

Use $M_1V_1 = M_2V_2$:
$M_{\text{OH}^-,2} = \frac{M_{\text{NaOH},1}V_{\text{NaOH},1}}{V_{\text{total}}} = \frac{0.250\ \text{M} \times 5.00\ \text{mL}}{20.0\ \text{mL}}$
$M_{\text{OH}^-,2} = 0.0625\ \text{M}$

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Step1: Define rate law for zero order

Rate $= k[\text{OH}^-]^0 = k$, slope $=-k$

Step2: Zero order slope (no $[\text{OH}^-]$ dependence)

$\text{Slope} = -0.003008$

Step3: Define rate law for first order

Rate $= k[\text{OH}^-]^1$, slope $=-k[\text{OH}^-]$

Step4: First order slope (3x $[\text{OH}^-]$)

$\text{Slope} = -0.003008 \times 3 = -0.009024$

Step5: Define rate law for second order

Rate $= k[\text{OH}^-]^2$, slope $=-k[\text{OH}^-]^2$

Step6: Second order slope (9x $[\text{OH}^-]^2$)

$\text{Slope} = -0.003008 \times 3^2 = -0.027072$

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Step1: Identify order for reactant Y

Only $1/[\text{Y}]$ vs time is linear: order = 2

Step2: Identify order for reactant Z

Only $\ln[\text{Z}]$ vs time is linear: order = 1

Step3: Write combined rate law

Rate $= k[\text{Y}]^2[\text{Z}]^1$

Answer:

1. Concentration of CV: $\boldsymbol{4.50 \times 10^{-5}\ \text{M}}$; Concentration of $\text{OH}^-$: $\boldsymbol{0.0625\ \text{M}}$
2. (a) Zero order: $\boldsymbol{-0.003008}$; (b) First order: $\boldsymbol{-0.009024}$; (c) Second order: $\boldsymbol{-0.027072}$
3. $\boldsymbol{\text{Rate} = k[\text{Y}]^2[\text{Z}]}$