Chemistry · The effect of temperature on the rate of reactions, Arrhenius theory, activation energy and its calculation, collision theory of bi-molecular gaseous reactions (no derivation)

Rate of disappearance of the reactant at two different temperature is given by }

Rate of disappearance of the reactant \( \boldsymbol{A} \) at two different temperature is given by \( \boldsymbol{A} \rightleftharpoons \boldsymbol{B} \) \( \frac{-\boldsymbol{d}[\boldsymbol{A}]}{\boldsymbol{d t}}=\left(\boldsymbol{2} \times \mathbf{1 0}^{-2} \boldsymbol{S}^{-1}\right)[\boldsymbol{A}]-\boldsymbol{4} \times \) \( \mathbf{1 0}^{-3} \boldsymbol{S}^{-\mathbf{1}}[\boldsymbol{B}], \boldsymbol{T}=\mathbf{3 0 0} \boldsymbol{K} \) \( \frac{-\boldsymbol{d}[\boldsymbol{A}]}{\boldsymbol{d t}}=\left(\boldsymbol{4} \times \mathbf{1 0}^{-2} \boldsymbol{S}^{-1}\right)[\boldsymbol{A}]-\mathbf{1 6} \times \) \( \mathbf{1 0}^{-\mathbf{4}}[\boldsymbol{B}], \boldsymbol{T}=\mathbf{4 0 0} \boldsymbol{K} \) Calculate heat of reaction in the given temperature range, when equilibrium is \( \operatorname{set} \) up.

  • A. \( 16.06 k J \)
  • B. \( 23.04 k J \)
  • C. \( 26.78 k J \)
  • D. \( 29.34 k J \)

Step-by-step solution

At equilibrium, -d[A]/dt = 0, so K = k_f/k_r. At 300 K, K1 = (2×10^{-2})/(4×10^{-3}) = 5. At 400 K, K2 = (4×10^{-2})/(16×10^{-4}) = 25. Using van't Hoff equation: ln(K2/K1) = ΔH/R (1/T1 - 1/T2). With R = 8.314 J/(mol·K), ΔH = 8.314 × ln(5) / (1/300 - 1/400) = 8.314 × 1.60944 × 1200 ≈ 16057 J/mol = 16.06 kJ/mol.
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