ANWSER
Question 1:
(a)
The five factors influencing enzyme-catalyzed reaction rates are:
1. Temperature: Increases rate up to an optimum, then denaturation occurs.
2. pH: Activity peaks at an optimal pH; deviations reduce activity.
3. Substrate Concentration: Rate increases until saturation (V_max).
4. Enzyme Concentration: Direct proportionality if substrate is abundant.
5. Inhibitors: Competitive/non-competitive inhibitors reduce activity.
(b) i.
For a first-order reaction:
\[ \ln\left(\frac{[A]}{[A]_0}\right) = -kt \]
At \( t_{1/2} \), \( [A] = \frac{[A]_0}{2} \):
\[ \ln\left(\frac{1}{2}\right) = -kt_{1/2} \implies t_{1/2} = \frac{\ln 2}{k} \]
No dependence on \([A]_0\).
ii.
Given \( k = 2.46 \times 10^{-4} \, \text{M}^{-1}\text{s}^{-1} \), this is a second-order reaction.
\[ t_{1/2} = \frac{1}{k[A]_0} = \frac{1}{(2.46 \times 10^{-4})(0.05)} = 81,300 \, \text{s} \, (22.6 \, \text{hr}) \]
(c)
i.
– General acid catalysis: All acids in solution contribute.
– Specific acid catalysis: Only \( \text{H}_3\text{O}^+ \) acts as catalyst.
ii.
– Homogeneous: Catalyst and reactants in same phase.
– Heterogeneous: Catalyst in different phase.
iii.
– Competitive inhibition: Inhibitor competes with substrate.
– Uncompetitive inhibition: Inhibitor binds only to enzyme-substrate complex.
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Question 2:
(a) i.
Solid phases (e.g., ice) have varied crystalline structures under different \( P \) and \( T \). Liquid/gas phases lack such order, leading to single phases.
ii.
Phase diagram of \( \text{D}_2\text{O} \):
– Axes: Temperature (\( T \)) vs. Pressure (\( P \)).
– Regions: Solid, liquid, vapor.
– Triple points (solid-liquid-vapor coexistence), critical point.
(b) i.
Clapeyron equation applies to:
1. Solid-liquid: \( \frac{dP}{dT} = \frac{\Delta H_{\text{fus}}}{T \Delta V} \).
2. Liquid-vapor: \( \frac{dP}{dT} = \frac{\Delta H_{\text{vap}}}{T \Delta V} \).
3. Solid-vapor: \( \frac{dP}{dT} = \frac{\Delta H_{\text{sub}}}{T \Delta V} \).
ii.
Given: \( \frac{dP}{dT} = 231 \, \text{atm/K} \), \( \Delta H_{\text{fus}} = 24.9 \, \text{cal/g} \), molar mass of phenol \( = 94.11 \, \text{g/mol} \).
Convert \( \Delta H_{\text{fus}} \):
\[ \Delta H = 24.9 \times 94.11 \times 4.184 = 9,800 \, \text{J/mol} \]
Using Clapeyron:
\[ \Delta V = \frac{\Delta H}{T \cdot \frac{dP}{dT}} = \frac{9,800}{314 \times (231 \times 101325)} \approx 1.1 \times 10^{-6} \, \text{m}^3/\text{mol} \]
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Question 3:
(a) i.
Colligative properties depend on solute particle count:
– Vapor pressure lowering, boiling point elevation, freezing point depression, osmotic pressure.
ii.
Assuming \( K_f \) for \( \text{CCl}_4 = 29.8 \, \text{K·kg/mol} \):
\[ \Delta T = K_f \cdot m \implies m = \frac{10.5}{29.8} = 0.352 \, \text{mol/kg} \]
Moles of solute: \( 0.352 \times 0.75 = 0.264 \, \text{mol} \).
Molar mass: \( \frac{100 \, \text{g}}{0.264 \, \text{mol}} = 379 \, \text{g/mol} \).
(b) i.
– Carnot engine: Reversible cycle with \( \eta = 1 – \frac{T_c}{T_h} \).
– Carnot refrigerator: COP \( = \frac{T_c}{T_h – T_c} \).
ii.
Efficiency between 4 K and 20 K:
\[ \eta = 1 – \frac{4}{20} = 0.8 \, (80\%) \]
For \( \eta = 0.8 \) and \( T_c = 300 \, \text{K} \):
\[ 0.8 = 1 – \frac{300}{T_h} \implies T_h = 1500 \, \text{K} \]
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Question 4:
(a)
Factors: Concentration, temperature, catalysts, ionic strength, solvent nature.
(b)
Collision theory:
\[ k = P \cdot Z \cdot e^{-E_a/(RT)} \]
For \( A + B \rightarrow P \), \( Z \) (collision frequency) \( \propto \sqrt{T} \), \( P \)=steric factor.
(c)
Using Eyring equation:
\[ k = \frac{k_B T}{h} e^{\Delta S^\ddagger/R} e^{-\Delta H^\ddagger/(RT)} \]
At \( T = 298 \, \text{K} \):
– \( E_a \approx \Delta H^\ddagger + RT = 185 + 2.48 = 187.48 \, \text{kJ/mol} \).
– Pre-exponential factor \( A \):
\[ A = \frac{(1.38 \times 10^{-23})(298)}{6.626 \times 10^{-34}} e^{15/8.314} \approx 1.2 \times 10^{14} \, \text{mol}^{-1}\text{dm}^3\text{s}^{-1} \]
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