The reaction 2HI → H2 + I2 is second order in [HI] and second order overall. The rate constant of the reaction at 700°C is 1.57 × 10−5 M −1s−1. Suppose you have a sample in which the concentration of HI is 0.75 M. What was the concentration of HI 8 hours earlier?
A) 0.45 M |
B) 0.75 M |
C) 2.3 M |
D) 1.9 M |
The reaction 2HI → H2 + I2 is second order in [HI] and second order overall....
Consider the second-order reaction: 2HI(g)→H2(g)+I2(g) Use the simulation to find the initial concentration [HI]0 and the rate constant k for the reaction. What will be the concentration of HI after t = 5.92×1010 s ([HI]t) for a reaction starting under the condition in the simulation?
The rate constant for the gaseous reaction H2(g) + I2(g) → 2HI(g) is 2.42 × 10−2/(M·s) at 400°C. Initially an equimolar sample of H2 and I2 is placed in a vessel at 400°C and the total pressure is 1690 mmHg. (a) What is the initial rate (M·min) of formation of HI? (__________) (b) What are the rate of formation of HI and the concentration of HI (in molarity) after 13.4 min? Rate of formation = (_______) (c) [HI] = (_______)...
Given the equilibrium reaction: 2HI(g) H2(g) + I2(g) A sample mixture of HI, H2, and 12, at equilibrium, was found to have [H2]- 1.4 x 102 Mand [HI 4.0 x 102 M. If Keq 1.0 x 10, calculate the molar concentration of I2 in the equilibrium mixture, Enter your answer in the provided box. ]= м
The equilibrium constant for the reaction: H2(g) + I2(g) <--> 2HI(g) is 54 at 700 K. A mixture of H2, I2 and HI, each at 0.020 M, was introduced into a container at 700 K. Which of the following is true? At equilibrium, [H2] = [I2] = [HI]. No net change occurs because the system is at equilibrium. The reaction proceeds to the left producing more H2(g) and I2(g). The reaction proceeds to the right producing more HI(g). At equilibrium,...
Calculate the equilibrium concentrations of H2, I2, and HI at 700 K if the initial concentrations are [H2] = 0.200 M and [I2] = 0.400 M. The equilibrium constant Kc for the reaction following reaction is 57.0 at 700 K. (Show Work) H2(g)+I2(g)<--- ---->2HI(g)
The gas phase reaction of hydrogen with iodine H2 + I2 ---------->2 HI is first order in H2 and first order in I2. Complete the rate law for this reaction in the box below. Use the form k[A]m[B]n... , where '1' is understood for m, n ... (don't enter 1) and concentrations taken to the zero power do not appear. Rate=__________ In an experiment to determine the rate law, the rate of the reaction was determined to be 5.37×10-22 Ms-1...
Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order in HI and the rate constant is 9.7×10−6M−1s−1. Part A What is the half-life (in days) of this reaction when the initial HI concentration is 0.120 M ? Express your answer using two significant figures. t1/2 t 1 / 2 = days Previous AnswersRequest Answer Incorrect; Try Again; 4 attempts remaining Part B How many days does it take for the concentration of HI...
The equilibrium constant for the following reaction is 1.80x102 at 698K 2HI(8)H2( ) +I2(g) If an equilibrium mixture of the three gases at 698K contains 2.35x102 M HI(g) and 2.63x10-2 M H, what is the equilibrium concentration of I? Submit Answer Retry Entire Group 9 more group attempts remaining
The reaction H2(g) + I2(g) → 2 HI (g) is first order in both hydrogen and iodine. It is therefore referred to as second order overall. Its rate constant for the formation of HI (g) at 400 ◦C is 2.34 × 10−2 · lit · mol−1 · sec−1 and its activation energy is 150 KJ/mol. Use the rate law to estimate how long it takes to form 0.1 mole of HI(g) if I start by putting 2 moles of H2...
For the reaction: 2HI(g) ↔ H2(g) + I2(g) Keq = 0.016 Initially a container contains 0.60 M HI, 0.038 M H2, and 0.15 M I2 at equilibrium. What is the new equilibrium concentration of H2, if the H2 concentration is increased by 0.276 M?