Two gaseous reactions have the following rate constants: X Y k₁ = 10⁸ e^ -15000 T P Q k₂ = 10⁶ e^ -x T If the rate constants of these two reactions become equal at a temperature of 1000 K, the value of x is _ _ _ _ . (Given: 10 = 2.3 )

Step-by-step solution

Given that k₁ = k₂ at T = 1000 K, we can equate the two expressions: 10⁸ e^ -15000 1000 = 10⁶ e^ -x 1000 Dividing both sides by 10⁶ gives: 10² e⁻¹⁵ = e^ -x 1000 Dividing both sides by e⁻¹⁵ gives: 10² = e^ 15000 - x 1000 Taking the natural logarithm on both sides: (10²) = 15000 - x 1000 2 10 = 15000 - x 1000 Substituting the given value of 10 = 2.3 : 2 2.3 = 15000 - x 1000 4.6 = 15000 - x 1000 4600 = 15000 - x x = 15000 - 4600 = 10400 Answer: 10400