Question 35 of 37intermediate🔧 ApplyNumerical3 marks

The decomposition of A into product has value of kk as 4.5×1034.5 \times 10^3 s−1^{-1} at 10∘C10^\circ\text{C} and energy of activation 60 kJ mol−1^{-1}. At what temperature would kk be 1.5×1041.5 \times 10^4 s−1^{-1}?

Correct Answer

297 K

Exercise: EXERCISES | Q: 3.28 | (Page 28)
For More Understanding

Explanation

To find the unknown temperature, we apply the Arrhenius equation which relates rate constants at two different temperatures. The formula log⁡k2k1=Ea2.303R[1T1−1T2]\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} [\frac{1}{T_1} - \frac{1}{T_2}] is derived from the context (specifically the section preceding Example 3.9). We must ensure all units are consistent: temperature in Kelvin and activation energy in Joules per mole. Solving the equation algebraically for T2T_2 yields the final temperature.

Solution Steps

  1. Step 1: Identify given data and convert units.

    k1=4.5×103 s−1k_1 = 4.5 \times 10^3 \text{ s}^{-1} T1=10∘C=10+273=283 KT_1 = 10^\circ\text{C} = 10 + 273 = 283 \text{ K} k2=1.5×104 s−1k_2 = 1.5 \times 10^4 \text{ s}^{-1} Ea=60 kJ mol−1=60000 J mol−1E_a = 60 \text{ kJ mol}^{-1} = 60000 \text{ J mol}^{-1} R=8.314 J K−1 mol−1R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}

  2. Step 2: Write the Arrhenius equation. Using the formula from the context:

    log⁡k2k1=Ea2.303R[1T1−1T2]\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]

  3. Step 3: Substitute the values into the equation.

    log⁡1.5×1044.5×103=600002.303×8.314[1283−1T2]\log \frac{1.5 \times 10^4}{4.5 \times 10^3} = \frac{60000}{2.303 \times 8.314} \left[ \frac{1}{283} - \frac{1}{T_2} \right]

  4. Step 4: Simplify and solve for T2T_2.

    log⁡103=6000019.147[1283−1T2]\log \frac{10}{3} = \frac{60000}{19.147} \left[ \frac{1}{283} - \frac{1}{T_2} \right] 0.5228=3133.57[0.003533−1T2]0.5228 = 3133.57 \left[ 0.003533 - \frac{1}{T_2} \right] 0.52283133.57=0.003533−1T2\frac{0.5228}{3133.57} = 0.003533 - \frac{1}{T_2} 0.0001668=0.003533−1T20.0001668 = 0.003533 - \frac{1}{T_2} 1T2=0.003533−0.0001668\frac{1}{T_2} = 0.003533 - 0.0001668 1T2=0.003366\frac{1}{T_2} = 0.003366 T2=10.003366≈297.03 KT_2 = \frac{1}{0.003366} \approx 297.03 \text{ K}

  5. Step 5: State the final answer. The temperature T2T_2 would be approximately 297 K.

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