The rate constants for two gaseous reactions are given as: Reaction 1: k₁ = 10¹⁵ e^ -10000 T Reaction 2: k₂ = 10¹¹ e^ -2000 T At a certain temperature T , the rate constants of the two reactions become equal ( k₁ = k₂ = k ). The value of ₁₀ k at this temperature is _ _ _ _ .

Step-by-step solution

Equating the two rate constants: 10¹⁵ e^ -10000 T = 10¹¹ e^ -2000 T Dividing both sides by 10¹¹ and by e^ -10000 T : 10⁴ = e^ 8000 T Taking the natural logarithm on both sides: (10⁴) = 8000 T 4 10 = 8000 T 1 T = 4 10 8000 = 10 2000 Now substitute this expression for 1 T back into the equation for k₂ (or k₁ ) to find the value of k : k = 10¹¹ e^ -2000 ( 10 2000 ) k = 10¹¹ e^ - 10 Since e^ - 10 = e^ (10⁻¹) = 10⁻¹ : k = 10¹¹ 10⁻¹ = 10¹⁰ Taking the base-10 logarithm: ₁₀ k = ₁₀(10¹⁰) = 10 Answer: 10