X+Y-> Products X Y Rate0.1 0.1 X0.2 0.2 4x0.4 0.2 16xWhat is the rate equation for the above reaction?
Answer (1)
This is a chemical kinetics problem: where X + Y → Products.[tex]\left[\begin{array}{ccc}[X](M)&[Y](M)&Rate(M/s)\\0.1&0.1&x\\0.2&0.2&4x\\0.4&0.2&16x\end{array}\right][/tex]The general rate law for this reaction is [tex]Rate = k[X]^m[Y]^n[/tex].k = rate constantm = order with respect to Xn = order with respect to YFind the order with respect to Y.Compare experiments 2 and 3 (where [X] changes but [Y] stays constant).From experiment 2 to 3doubles: 0.2 → 0.4[Y] remains constant: 0.2 → 0.2Rate changes: 4x → 16x (increases by factor of 4)Since [Y] is constant, we can write [tex]Rate_{3}/Rate_{2} = ([X]_{3} /[X]_{2} )^m[/tex].[tex]16x/4x = (0.4/0.2)^m\\4 = (2)^m\\2^{2} = 2^m\\m = 2[/tex]Find the order with respect to Y.Compare experiments 1 and 2 (where both [X] and [Y] change proportionally).From experiment 1 to 2doubles: 0.1 → 0.2[Y] doubles: 0.1 → 0.2Rate changes: x → 4x (increases by factor of 4)[tex]Rate = k[X]^{2} [Y]^n\\Rate_{2}/Rate_{1} = [([X]_{2}/[X]_{1})^{2}] [([Y]_{2}/[Y]_{1})^n]\\4x/x = [(0.2/0.1)^{2}][(0.2/0.1)^n]\\4 = [(2)^{2}][(2)^n]\\4 = (4)(2^n)\\1 = 2^n\\2^{0} = 2^n\\n = 0[/tex]Determine the rate law.Rate = k[X]²[Y]⁰ = k[X]²ReactionSecond order with respect to XZero order with respect to YSecond order overallCalculate the rate constant.Use experiment 1 data.Rate = k[X]²x = k(0.1)²x = k(0.01)k = 100x M⁻¹s⁻¹Verify with all experiments:Experiment 1: Rate = 100x × (0.1)² = 100x × 0.01 = x ✓Experiment 2: Rate = 100x × (0.2)² = 100x × 0.04 = 4x ✓Experiment 3: Rate = 100x × (0.4)² = 100x × 0.16 = 16x ✓Rate Law: Rate = k[X]²Rate Constant: k = 100x M⁻¹s⁻¹Order: 2nd order in X, 0th order in Y, 2nd order overall
