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ATTEMPT ANY TWO (2) QUESTIONS

Chemistry

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Here are the solutions to the questions:

1a) Define the following:

i) Rate of reaction:
The rate of reaction is the change in the concentration of a reactant or product per unit time. It measures how fast reactants are consumed or products are formed.

ii) Half-Life:
Half-life (t1/2t_{1/2}) is the time required for the concentration of a reactant to decrease to half of its initial concentration.

iii) Activation energy:
Activation energy (EaE_a) is the minimum amount of energy that reactant molecules must possess to undergo a chemical reaction. It represents the energy barrier that must be overcome for a reaction to occur.

1b)

i) Write the rate equation for the reaction 2A→B+C2A \to B + C if the order of the reaction is zero.
For a zero-order reaction, the rate is independent of the reactant concentration.

Rate=k[A]0=k\text{Rate} = k[A]^0 = k

The rate equation is Rate=k\boxed{\text{Rate} = k}.

ii) For a chemical reaction A→BA \to B, it is found that the rate of reaction doubles when the concentration of AA is increased four times. The order of the reaction is:
Let the rate law be Rate=k[A]n\text{Rate} = k[A]^n.
If [A][A] is increased by 4 times, the rate doubles.

2×Rate1=k(4[A]1)n2 \times \text{Rate}_1 = k(4[A]_1)^n 2=4n2 = 4^n

To solve for nn, we can write 4n=(22)n=22n4^n = (2^2)^n = 2^{2n}.

21=22n2^1 = 2^{2n} 1=2n1 = 2n n=12n = \frac{1}{2}

The order of the reaction is 12\boxed{\frac{1}{2}}.

iii) For a reaction R→PR \to P, half-life (t1/2t_{1/2}) is observed to be independent of the initial concentration of reactants. What is the order of reaction?
For a first-order reaction, the half-life is independent of the initial concentration (t1/2=ln⁡2kt_{1/2} = \frac{\ln 2}{k}). For other orders, it depends on the initial concentration.
The order of reaction is First order\boxed{\text{First order}}.

1c)
CaCO3\text{CaCO}_3 pieces are placed in a flask and HCl\text{HCl} is added. Immediately a balloon is placed over the top of the flask and the balloon starts to inflate.

i) Explain why the balloon inflates.
The reaction between calcium carbonate (CaCO3\text{CaCO}_3) and hydrochloric acid (HCl\text{HCl}) produces carbon dioxide gas (CO2\text{CO}_2). This gas accumulates inside the flask, increasing the pressure and causing the balloon to inflate.

CaCO3(s)+2HCl(aq)→CaCl2(aq)+H2O(l)+CO2(g)\text{CaCO}_3(s) + 2\text{HCl}(aq) \to \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

ii) If the same mass of CaCO3\text{CaCO}_3 in a powdered form is used, the balloon inflates faster. Explain why using collision theory.
Using powdered CaCO3\text{CaCO}_3 increases the surface area available for the reaction. According to collision theory, a larger surface area leads to more frequent effective collisions between the reactant particles (CaCO3\text{CaCO}_3 and HCl\text{HCl}), thus increasing the rate of reaction and producing CO2\text{CO}_2 gas faster, which inflates the balloon more quickly.

iii) Suggest two other ways to make the balloon inflate faster.

  • Increase the concentration of HCl\text{HCl}: A higher concentration of HCl\text{HCl} means more reactant particles per unit volume, leading to more frequent effective collisions and a faster reaction rate.
  • Increase the temperature: Raising the temperature increases the kinetic energy of the reactant molecules, causing them to move faster and collide more frequently and with greater energy, leading to a higher proportion of effective collisions and a faster reaction rate.

1d) Consider the following reactions. Predict and arrange the reactions as they occur from the most rapidly at room conditions to the slowest.

The general principles for reaction rates are:

  • Ionic reactions in solution are typically very fast.
  • Reactions involving gases or complex molecules, especially those requiring significant bond breaking and forming, tend to be slower.
  • Decomposition reactions can vary but are often slow without catalysts.

Let's analyze each reaction:

i) C6H12O6+O2→2CO2+3H2O\text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \to 2\text{CO}_2 + 3\text{H}_2\text{O} (Combustion/oxidation of glucose) - This is a complex organic reaction involving many bond rearrangements. It is generally slow at room temperature without initiation.

ii) Fe2++O2→Fe2O3\text{Fe}^{2+} + \text{O}_2 \to \text{Fe}_2\text{O}_3 (Oxidation of iron(II) to iron(III) oxide) - This is a redox reaction involving a gas (O2\text{O}_2) and ions. While faster than complex organic reactions, it's typically slower than simple ionic reactions in solution.

iii) H2O2→H2O+O2\text{H}_2\text{O}_2 \to \text{H}_2\text{O} + \text{O}_2 (Decomposition of hydrogen peroxide) - This reaction is relatively slow at room temperature without a catalyst.

iv) 2Fe3++Sn2+→2Fe2++Sn4+2\text{Fe}^{3+} + \text{Sn}^{2+} \to 2\text{Fe}^{2+} + \text{Sn}^{4+} (Redox reaction between ions) - This is a reaction between ions in solution, which typically occurs very rapidly due to minimal bond breaking and strong electrostatic attractions.

Arranging from most rapidly to slowest:

  1. iv) 2Fe3++Sn2+→2Fe2++Sn4+2\text{Fe}^{3+} + \text{Sn}^{2+} \to 2\text{Fe}^{2+} + \text{Sn}^{4+} (Fastest - ionic reaction)
  2. ii) Fe2++O2→Fe2O3\text{Fe}^{2+} + \text{O}_2 \to \text{Fe}_2\text{O}_3 (Moderately fast - redox, but involves gas)
  3. iii) H2O2→H2O+O2\text{H}_2\text{O}_2 \to \text{H}_2\text{O} + \text{O}_2 (Slow - decomposition without catalyst)
  4. i) C6H12O6+O2→2CO2+3H2O\text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \to 2\text{CO}_2 + 3\text{H}_2\text{O} (Slowest - complex organic reaction)

The order from most rapidly to slowest is:
iv) > ii) > iii) > i)\boxed{\text{iv) } > \text{ ii) } > \text{ iii) } > \text{ i)}}

2a) A substance dissociates through a first-order reaction with a half-life of 32 min. If the initial concentration of the substance is 2.7×10112.7 \times 10^{11} molecules/L, what will be the concentration 2.24 h later?

  1. Calculate the rate constant (kk) from the half-life (t1/2t_{1/2}).

    For a first-order reaction:

    t1/2=ln⁡2kt_{1/2} = \frac{\ln 2}{k} k=ln⁡2t1/2=0.69332 min=0.021656 min−1k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{32 \text{ min}} = 0.021656 \text{ min}^{-1}
  2. Convert the total time to minutes.

    Time (t)=2.24 h×60 min/h=134.4 min\text{Time } (t) = 2.24 \text{ h} \times 60 \text{ min/h} = 134.4 \text{ min}
  3. Use the integrated rate law for a first-order reaction to find the concentration at time tt.

    ln⁡[A]t=−kt+ln⁡[A]0\ln[A]_t = -kt + \ln[A]_0 ln⁡[A]t=−(0.021656 min−1)(134.4 min)+ln⁡(2.7×1011 molecules/L)\ln[A]_t = -(0.021656 \text{ min}^{-1})(134.4 \text{ min}) + \ln(2.7 \times 10^{11} \text{ molecules/L}) ln⁡[A]t=−2.907+26.319\ln[A]_t = -2.907 + 26.319 ln⁡[A]t=23.412\ln[A]_t = 23.412 [A]t=e23.412[A]_t = e^{23.412} [A]t=1.47×1010 molecules/L[A]_t = 1.47 \times 10^{10} \text{ molecules/L}

    The concentration 2.24 h later will be 1.47×1010 molecules/L\boxed{1.47 \times 10^{10} \text{ molecules/L}}.

    2b)

    i) Express the rate of the following reaction in terms of the formation of ammonia:
    N2(g)+3H2(g)→2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)
    The rate of reaction can be expressed in terms of the change in concentration of reactants and products, divided by their stoichiometric coefficients. For the formation of ammonia:

    Rate=+12d[NH3]dt\text{Rate} = +\frac{1}{2}\frac{d[\text{NH}_3]}{dt}

    The rate of reaction in terms of the formation of ammonia is Rate=+12d[NH3]dt\boxed{\text{Rate} = +\frac{1}{2}\frac{d[\text{NH}_3]}{dt}}.

    ii) For the reaction 3A+2B→C+2D3A + 2B \to C + 2D, reactant AA is found to disappear at the rate of 4.6×10−3 Ms−14.6 \times 10^{-3} \text{ Ms}^{-1}.

    xiii) What is the rate of disappearance of the reactant BB?
    The relationship between the rates of disappearance of AA and BB is given by their stoichiometric coefficients:

    −13d[A]dt=−12d[B]dt-\frac{1}{3}\frac{d[A]}{dt} = -\frac{1}{2}\frac{d[B]}{dt}

    We are given −d[A]dt=4.6×10−3 Ms−1-\frac{d[A]}{dt} = 4.6 \times 10^{-3} \text{ Ms}^{-1}.

    −12d[B]dt=13×(4.6×10−3 Ms−1)-\frac{1}{2}\frac{d[B]}{dt} = \frac{1}{3} \times (4.6 \times 10^{-3} \text{ Ms}^{-1}) −d[B]dt=23×(4.6×10−3 Ms−1)-\frac{d[B]}{dt} = \frac{2}{3} \times (4.6 \times 10^{-3} \text{ Ms}^{-1}) −d[B]dt=3.066...×10−3 Ms−1-\frac{d[B]}{dt} = 3.066... \times 10^{-3} \text{ Ms}^{-1}

    The rate of disappearance of reactant BB is 3.07×10−3 Ms−1\boxed{3.07 \times 10^{-3} \text{ Ms}^{-1}}.

    xiv) What is the rate of appearance for the product DD?
    The relationship between the rate of disappearance of AA and the rate of appearance of DD is:

    −13d[A]dt=+12d[D]dt-\frac{1}{3}\frac{d[A]}{dt} = +\frac{1}{2}\frac{d[D]}{dt}

    We are given −d[A]dt=4.6×10−3 Ms−1-\frac{d[A]}{dt} = 4.6 \times 10^{-3} \text{ Ms}^{-1}.

    +12d[D]dt=13×(4.6×10−3 Ms−1)+\frac{1}{2}\frac{d[D]}{dt} = \frac{1}{3} \times (4.6 \times 10^{-3} \text{ Ms}^{-1}) +d[D]dt=23×(4.6×10−3 Ms−1)+\frac{d[D]}{dt} = \frac{2}{3} \times (4.6 \times 10^{-3} \text{ Ms}^{-1}) +d[D]dt=3.066...×10−3 Ms−1+\frac{d[D]}{dt} = 3.066... \times 10^{-3} \text{ Ms}^{-1}

    The rate of appearance for product DD is 3.07×10−3 Ms−1\boxed{3.07 \times 10^{-3} \text{ Ms}^{-1}}.

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