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Question 24 4 pts Molecular iodine, 12 (g), dissociates into iodine atoms at 652 K, with...
Molecular iodine, I2 (g), dissociates into iodine atoms at 652 K, with a first order rate constant of 0.452 s-1. a. What is the half life for this reaction? [ Select ] ["1.53 s", "0.31 s", "0.079 s", "0.39 s"] b. If you start with 0.35 M I2 at this temperature, how long will it take for the concentration of I2 to reach...
Molecular iodine, I2(g), dissociates into iodine atoms at 625 K with a first-order rate constant of 0.271 s-1. (a) What is the half-life for this reaction? _______ s (b) If you start with 0.051 M I2 at this temperature, how much will remain after 5.30 s assuming that the iodine atoms do not recombine to form I2? _______ M
The dissociation of molecular iodine into iodine atoms is represented as I2(g) ⇌ 2I(g) At 1000 K, the equilibrium constant Kc for the reaction is 3.80 × 10−5. Suppose you start with 0.0458 mol of I2 in a 2.32−L flask at 1000 K. What are the concentrations of the gases at equilibrium? What is the equilibrium concentration of I2? __M What is the equilibrium concentration of I? __M
21.Be sure to answer all parts. The dissociation of molecular iodine into iodine atoms is represented as I2(g) ⇌ 2I(g) At 1000 K, the equilibrium constant Kc for the reaction is 3.80 ×10−5. Suppose you start with 0.0461 mol of I2 in a 2.27−L flask at 1000 K. What are the concentrations of the gases at equilibrium? What is the equilibrium concentration of I2? What is the equilibrium concentration of I?
1)Consider the following equilibrium at 972 K for the dissociation of molecular iodine into atoms of iodine. I2(g) ----> 2 I(g); Kc = 1.40 ? 10?3 Suppose this reaction is initiated in a 3.4 L container with 0.067 mol I2 at 972 K. Calculate the concentrations of I2 and I at equilibrium. * I got I2= 0.0135M and I= 0.0124 is this right?* 2)Consider the following equilibrium. NH3(aq) + H2O(l) -----> NH4+(aq) + OH ?(aq) What will happen to the...
Need help with Part B KAssignment 16 Chap 14: Integrated Rate Law and Half Life Problem 14.42 - Enhanced - with Feedback 5 of 7 Review I Constants I Periodic Table Molecular iodine, I2 (g), dissociates into iodine atoms at 625 K with a first-order rate constant of What is the half-life for this reaction? -1 0.271 S Express the half-life in seconds to three significant figures. You may want to reference (Pages 582-587) Section 14.4 while completing this problem....
9. The equilibrium constant(K.) for the gaseous dissociation of Iodine molecules to lodine atoms (shown below) 12(8) 21(8) is 3.76 x 10-3 at 1000K. If 0.15 mole of 12(g) is placed in a 12.3-L flask at 1000 K, what are the concentrations of 12(g) and I(g) when the reaction comes to equilibrium?
1. A reaction is second order in[A] and second-order in [B]: Rate,=K[A]^2[B]^2. what are the units of k for this reaction? If the concentration of air decreases by a factor of 2 and the concentration of b increases by a factor of 5 what happens to the rate? 2. for the forward reaction 2NO+Cl2=>2NOCl. determine the rate(m/s)for experiment #4 given [NO]°(M)=0.40M and [Cl2]°z(M)=0?20M. Rate? 3.The following data were collected over time for the forward reaction 2NO2=>2NO+O2 ( 1/[NO2]=100 at 0...
The rate law for the reaction 3A(g) → C(g) is rate-kAI2 where k = 4.36 x 10-2 Mhrl. The initial concentration of A is 0.250 M. a) How long will it take for [A] to reach 1/6 of its initial concentration b) What is the half-life of this reaction? c) What is the concentration of A after 10.0 hr?
QUESTION 24 4 Consider the following reversible reaction: H2(g) + 12(9) 2 HI(g) K = 64.0 at a particular temperature. A reaction mixture initially contains 0.945 M H2 and 0.945 M 12. Determine the equilibrium concentration of HL Click Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answer 0 1 2 v