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- What would be the rate law for the mechanism below? Step 1 NO(g) + Cl2(g) ↔ NOCl2(g) slow Step 2 NOCl2(g) + NO(g) → 2 NOCl(g) fast kobs[NOCl2][NO] kobs[NO][Cl2] kobs[Cl2][NO]2 kobs[NO]2This table provides rate data for the hypothetical rxn, 2A + B ® C. What is its rate law? Rate = k[A][B] Rate = k[A]2[B] Rate = k[A][B]2 Rate = k[A]2[B]2 Rate = k[B]2Q3. Find the value of the rate constant at 600 K
- What is the molecularity of the rate limiting step for the mechanism below? Step 1: H2O2 (aq) + I–(aq) → H2O(l) + OI–(aq) (slow)Step 2: H2O2 (aq) + OI–(aq) → H2O(l) + O2 (g) + I–(aq) (fast)Is the proposed mechanism below valid or invalid? Justify your answer.Overall reaction: 2NO2(g) + F2(g) → 2NO2F(g) ________________________________________________Step 1: NO2(g) + F2(g) →NO2F(g) + F(g) slow Step 2: F(g) + NO2(g) → NO2F(g) fastrate=k[CH3Cl][Cl2]^(1/2) Calculate the value of the rate constant, k.
- Rate information was obtained for the following reaction at 25°C and 33°C; Cr(H2O)6 3+ + SCN¯ ---> Cr(H2O)5NCS2+ + H2O(l) Initial Rate [Cr(H2O)6 3+] [SCN-] @ 25°C 2.0x10-11 1.0x10-4 0.10 2.0x10-10 1.0x10-3 0.10 9.0x10-10 2.0x10-3 0.15 2.4x10-9 3.0x10-3 0.20 @ 33°C 1.4x10-10 1.0x10-4 0.20 Can a 4th order reaction occur?Rate information was obtained for the following reaction at 25°C and 33°C; Cr(H2O)6 3+ + SCN¯ ---> Cr(H2O)5NCS2+ + H2O(l) Initial Rate [Cr(H2O)6 3+] [SCN-] @ 25°C 2.0x10-11 1.0x10-4 0.10 2.0x10-10 1.0x10-3 0.10 9.0x10-10 2.0x10-3 0.15 2.4x10-9 3.0x10-3 0.20 @ 33°C 1.4x10-10 1.0x10-4 0.20 a) Write a rate law consistent with the experimental data. b) What is the value of the rate constant at 25°C? c) What is the value of the rate constant at 33°C? d) What is the activation energy for this reaction? e) Can a 4th order reaction occur?How is ΔG related to the forward (k1) and reverse(k-1) rates of a reversible first order reaction? ΔG = -RTln(k1/k-1) (k-1/k1) = e(-ΔG/RT) k1/k-1) = -e(ΔG/RT) ΔG = RTln(k1/k-1)
- rate law is r=k[ClO2][OH-]Determine the rate law for the overall reaction below using the mechanisms listed: F2 + CHF3 → HF + CF4 (overall) Step #1: F2 ⇌ 2 F (fast) Step #2: F + CHF3 → HF + CF3 (slow) Step #3: F + CF3 → CF4 (fast) rate = k [F2]½[CHF3] rate = k [F]2 rate = k [F][CHF3] rate = k [F2]½ rate = k [F2][CHF3]What would be the rate law for the mechanism below? Step 1 I2 → 2 I fast Step 2 2 I + H2 ↔ 2 HI slow