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A) State the Arrhenius equation and explain its application in chemical kinetics. b) Explain the use of simultaneous equations in chemistry with an example.

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QUESTION ONE

a) The Arrhenius equation is:

k=Ae−Ea/RTk = A e^{-E_a / RT}

where kk is the rate constant, AA is the pre-exponential factor (frequency factor), EaE_a is the activation energy, RR is the ideal gas constant, and TT is the absolute temperature.

Its application in chemical kinetics is to describe the temperature dependence of reaction rates. It shows that as temperature increases, the rate constant (and thus the reaction rate) increases exponentially due to a larger fraction of molecules possessing energy equal to or greater than the activation energy. It also allows for the determination of activation energy from experimental rate constants at different temperatures.

b) Simultaneous equations in chemistry are used to solve problems involving multiple unknown concentrations or quantities that are related by several independent equations. These equations often arise in equilibrium systems, mixtures, or analytical determinations where multiple species or reactions are present.

Example 1: Acid-Base Equilibria
In a solution containing a polyprotic acid (e.g., H2CO3\text{H}_2\text{CO}_3), there are multiple dissociation steps, each with its own equilibrium constant (Ka1,Ka2K_{a1}, K_{a2}). To find the concentrations of all species (H2CO3\text{H}_2\text{CO}_3, HCO3−\text{HCO}_3^-, CO32−\text{CO}_3^{2-}, H+\text{H}^+, OH−\text{OH}^-), a set of simultaneous equations (mass balance, charge balance, and equilibrium constant expressions) must be solved.

Example 2: Mixture Analysis
If a mixture contains two components, A and B, and two different analytical methods are used, each providing a total measurement related to the concentrations of A and B, then two simultaneous equations can be set up to determine the individual concentrations of A and B. For instance, if method 1 measures CA+CB=Total1C_A + C_B = \text{Total}_1 and method 2 measures 2CA+CB=Total22C_A + C_B = \text{Total}_2, these can be solved simultaneously.

c)

  1. Rearrange the equation v=u+atv = u + at to make aa the subject.

    v−u=atv - u = at a=v−uta = \frac{v - u}{t}

  2. Convert the time from minutes to seconds.

    t=4 minutes×60 seconds/minute=240 secondst = 4 \text{ minutes} \times 60 \text{ seconds/minute} = 240 \text{ seconds}

  3. Substitute the given values (u=0u=0 m/s, v=20v=20 m/s, t=240t=240 s) into the rearranged equation to find the acceleration aa.

    a=20 m/s−0 m/s240 sa = \frac{20 \text{ m/s} - 0 \text{ m/s}}{240 \text{ s}} a=20240 m/s2a = \frac{20}{240} \text{ m/s}^2 a=112 m/s2a = \frac{1}{12} \text{ m/s}^2 a≈0.0833 m/s2a \approx 0.0833 \text{ m/s}^2

    The acceleration is 0.0833 m/s2\boxed{\text{0.0833 m/s}^2}.

    d)

  4. Calculate the half-life (t1/2t_{1/2}) for a first-order decomposition.

    For a first-order reaction, the half-life is given by:

    t1/2=ln⁡2kt_{1/2} = \frac{\ln 2}{k}

    Given k=2.5×10−3 s−1k = 2.5 \times 10^{-3} \text{ s}^{-1} and ln⁡2=0.693\ln 2 = 0.693:

    t1/2=0.6932.5×10−3 s−1t_{1/2} = \frac{0.693}{2.5 \times 10^{-3} \text{ s}^{-1}} t1/2=277.2 st_{1/2} = 277.2 \text{ s}

    The half-life is 277.2 s\boxed{\text{277.2 s}}.

  5. Calculate the time to reach 15% of the initial concentration.

    The integrated rate law for a first-order reaction is:

    ln⁡([A]t[A]0)=−kt\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt

    We want the time when [A]t=0.15[A]0[A]_t = 0.15 [A]_0.

    ln⁡(0.15[A]0[A]0)=−(2.5×10−3 s−1)t\ln\left(\frac{0.15 [A]_0}{[A]_0}\right) = - (2.5 \times 10^{-3} \text{ s}^{-1}) t ln⁡(0.15)=−(2.5×10−3 s−1)t\ln(0.15) = - (2.5 \times 10^{-3} \text{ s}^{-1}) t

  6. Solve for tt.

    ln⁡(0.15)≈−1.897\ln(0.15) \approx -1.897 −1.897=−(2.5×10−3 s−1)t-1.897 = - (2.5 \times 10^{-3} \text{ s}^{-1}) t t=−1.897−2.5×10−3 s−1t = \frac{-1.897}{-2.5 \times 10^{-3} \text{ s}^{-1}} t=758.8 st = 758.8 \text{ s}

    The time to reach 15% of the initial concentration is 758.8 s\boxed{\text{758.8 s}}.

    QUESTION TWO

    a)
    Given data points (xix_i): 35.8, 36.4, 36.1, 36.0, 36.7.
    Number of data points (nn) = 5.

  7. Calculate the mean (xˉ\bar{x}).

    xˉ=∑xin\bar{x} = \frac{\sum x_i}{n} xˉ=35.8+36.4+36.1+36.0+36.75\bar{x} = \frac{35.8 + 36.4 + 36.1 + 36.0 + 36.7}{5} xˉ=181.05\bar{x} = \frac{181.0}{5} xˉ=36.2 ppm\bar{x} = 36.2 \text{ ppm}

    The mean is 36.2 ppm\boxed{\text{36.2 ppm}}.

  8. Calculate the variance (s2s^2).

    The formula for sample variance is s2=∑(xi−xˉ)2n−1s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}.
    Calculate (xi−xˉ)2(x_i - \bar{x})^2 for each point:
    (35.8−36.2)2=(−0.4)2=0.16(35.8 - 36.2)^2 = (-0.4)^2 = 0.16
    (36.4−36.2)2=(0.2)2=0.04(36.4 - 36.2)^2 = (0.2)^2 = 0.04
    (36.1−36.2)2=(−0.1)2=0.01(36.1 - 36.2)^2 = (-0.1)^2 = 0.01
    (36.0−36.2)2=(−0.2)2=0.04(36.0 - 36.2)^2 = (-0.2)^2 = 0.04
    (36.7−36.2)2=(0.5)2=0.25(36.7 - 36.2)^2 = (0.5)^2 = 0.25
    Sum of squared differences: ∑(xi−xˉ)2=0.16+0.04+0.01+0.04+0.25=0.50\sum(x_i - \bar{x})^2 = 0.16 + 0.04 + 0.01 + 0.04 + 0.25 = 0.50.

    s2=0.505−1=0.504s^2 = \frac{0.50}{5-1} = \frac{0.50}{4} s2=0.125 ppm2s^2 = 0.125 \text{ ppm}^2

    The variance is 0.125 ppm2\boxed{\text{0.125 ppm}^2}.

  9. Calculate the sample standard deviation (ss).

    s=s2s = \sqrt{s^2} s=0.125 ppm2s = \sqrt{0.125 \text{ ppm}^2} s≈0.3536 ppms \approx 0.3536 \text{ ppm}

    The sample standard deviation is 0.354 ppm\boxed{\text{0.354 ppm}}.

    b)
    Given: Population standard deviation σ=0.020\sigma = 0.020 mL, average volume Vˉ=24.86\bar{V} = 24.86 mL, number of titrations n=3n = 3.
    We need to find the 95% confidence interval for the true average volume.
    For a 95% confidence interval, the z-score (zα/2z_{\alpha/2}) is 1.96.

  10. Calculate the standard error of the mean.

    Standard Error=σn\text{Standard Error} = \frac{\sigma}{\sqrt{n}} Standard Error=0.020 mL3\text{Standard Error} = \frac{0.020 \text{ mL}}{\sqrt{3}} Standard Error≈0.020 mL1.732\text{Standard Error} \approx \frac{0.020 \text{ mL}}{1.732} Standard Error≈0.011547 mL\text{Standard Error} \approx 0.011547 \text{ mL}

  11. Calculate the margin of error.

    Margin of Error=zα/2×Standard Error\text{Margin of Error} = z_{\alpha/2} \times \text{Standard Error} Margin of Error=1.96×0.011547 mL\text{Margin of Error} = 1.96 \times 0.011547 \text{ mL} Margin of Error≈0.02263 mL\text{Margin of Error} \approx 0.02263 \text{ mL}

  12. Construct the 95% confidence interval.

    Confidence Interval=Vˉ±Margin of Error\text{Confidence Interval} = \bar{V} \pm \text{Margin of Error} Confidence Interval=24.86 mL±0.02263 mL\text{Confidence Interval} = 24.86 \text{ mL} \pm 0.02263 \text{ mL}

    Lower bound: 24.86−0.02263=24.83737 mL24.86 - 0.02263 = 24.83737 \text{ mL}
    Upper bound: 24.86+0.02263=24.88263 mL24.86 + 0.02263 = 24.88263 \text{ mL}
    Rounding to two decimal places (consistent with σ\sigma):
    The 95% confidence interval is 24.84 mL to 24.88 mL\boxed{\text{24.84 mL to 24.88 mL}}.

    c)
    Given: Formic acid (HCOOH) is a weak acid.
    Ka=1.8×10−4K_a = 1.8 \times 10^{-4}.
    Equation: Ka=[H+]2M−[H+]K_a = \frac{[\text{H}^+]^2}{M - [\text{H}^+]}.
    Initial moles of formic acid = 0.1 moles. Assuming the volume is 1 L, then the initial concentration M=0.1M = 0.1 M.
    Let [H+]=x[\text{H}^+] = x.

  13. Substitute

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