The reaction 2HI --> H2 + I2 is second order in [HI] and second order overall....

The reaction 2HI --> H₂ + I₂ is second order in [HI] and second order overall. The rate constant of the reaction at 700 degrees celsius is 1.57 x 10^-5 M^-1 s^-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?

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queen_honey_blossom answered 2 years ago

The gas phase reaction H2 + I2 --> 2HI is second order. Its rate constant at...

The gas phase reaction H2 + I2 --> 2HI is second order. Its rate constant at 400 C is 0.0243 dm3/mol s. Calculate delta H, delta S, delta G and the pre-exponential at this temperature. Assume that delta H is constant over this temperature range.

H2 + I2 yields 2HI At the start of the reaction 1 mole of I2 and...

H2 + I2 yields 2HI At the start of the reaction 1 mole of I2 and 1 mole of H2 are put into a 10 liter container. There is no HI present. At the reaction temperature Keq is 64 What is the equilibrium concentrations for H2, I2 and HI?

2HI --> H2 + I2 a mixture of 0.500 M HI, 0.100 M H2, and 0.100...

2HI --> H2 + I2 a mixture of 0.500 M HI, 0.100 M H2, and 0.100 M I2 is placed in a reaction vessel and allowed to come to equilibrium at a temperature of 745 K. The value of the equilibrium constant at that temperature is 0.0200. What is the concentration of HI at equilibrium?

For the following reaction: 2HI (g) ----> H2(g) + I2(g), the rate law is rate= k[HI]^2....

For the following reaction: 2HI (g) ----> H2(g) + I2(g), the rate law is rate= k[HI]^2. You are starting with 1.0M HI. How long will the first half-life reaction be? (The professor adds: Note that since you don't know the value for the rate constant k, just leave it in your answers.)

At 500 °C, hydrogen iodide decomposes according to 2HI(g)↽−−⇀H2(g)+I2(g)2HI(g)↽−−⇀H2(g)+I2(g) For HI(g)HI(g) heated to 500 °C in...

At 500 °C, hydrogen iodide decomposes according to 2HI(g)↽−−⇀H2(g)+I2(g)2HI(g)↽−−⇀H2(g)+I2(g) For HI(g)HI(g) heated to 500 °C in a 1.00 L reaction vessel, chemical analysis determined these concentrations at equilibrium: [H2]=0.400 M[H2]=0.400 M , [I2]=0.400 M[I2]=0.400 M , and [HI]=3.38 M[HI]=3.38 M . If an additional 1.00 mol of HI(g)HI(g) is introduced into the reaction vessel, what are the equilibrium concentrations after the new equilibrium has been reached? [HI]= [H2]= [I2]= Please help and show work!

For the following reaction: H2(g) + I2(g)<--->2HI(g) ; Kc=49.5 The beginning concentrations are: H2: 0.10M, I2:...

For the following reaction: H2(g) + I2(g)<--->2HI(g) ; Kc=49.5 The beginning concentrations are: H2: 0.10M, I2: 0.050M, and HI:0 Assume this reaction is not at equilibrium. When it does reach equilibrium, what will the concentrations be?

1. Consider the reaction for the production of HI from H2 and I2. H2 (g)+...

1. Consider the reaction for the production of HI from H2 and I2. H2 (g)+ I2(g) D 2HI(g) a) What is the initial concentration of HI? b) What is the equilibrium concentration of HI? c) How is the change in concentration of HI calculated? Explain: d) How are the changes in concentration for H2 and I2 calculated? Explain e) How are the equilibrium concentrations of H2 and I2 calculated? Explain

For the reaction H2(g) + I2(g) ↔ 2HI the value of the equilibrium constant is 25....

For the reaction H2(g) + I2(g) ↔ 2HI the value of the equilibrium constant is 25. Starting with 1.00mol of each reactant in a 10.L vessel, how many moles of HI will be present at equilibrium?

Consider the gas phase reaction H2 + I2 = 2HI at equilibrium at T= 870 K...

Consider the gas phase reaction H2 + I2 = 2HI at equilibrium at T= 870 K and p = 1 bar. If ∆G˚ (870K, 1 bar) = - 48.96 kJ, determine K, ξ and the equilibrium amounts of H2, I2 and HI (in mol) if the intial composition was n˚(H2) = 0.300 mol, n˚(I2) = 0.400 mol, and n˚(HI) = 0.200 mol

Calculate ΔGº at 25ºC for the reaction H2 + I2 D 2HI. Starting with 1.0 atm...

Calculate ΔGº at 25ºC for the reaction H2 + I2 D 2HI. Starting with 1.0 atm H2 and 1.0 atm I2, What is ΔG when: 0.1%, 1%, 10%, 50%, 90%, 99%, and 99.9% has reacted.

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