4.00 mols of H2 and 3 mols of I2 are placed in an evacuated 5.00L flask and then heated to 800K. The system is allowed to reach equilibrium. what will be the equilibrium concentration of each species? 2HI(g) <-----> H2(g) + I2(g) Kc= 0.016 @ 800k
Volume of Flask = 5.00 L
Moles of H2 = 4.00 mol
[H2] = moles/ Volume = 4mol/ 5 L = 0.8 M
[I2] = 3 mol/ 5 L = 0.6 M
Since In beaker we have only products. Thus reaction will shift backward as:
| 2HI(g) | <--> | H2(g)+ | I2(g) | |
| I(M) | 0 | 0.8 | 0.6 | |
| C(M) | +2x | -x | -x | |
| E(M) | 2x | 0.8-x | 0.6-x |
Kc = [H2][I2]/[HI]2
0.016 = (0.8-x)(0.6-x) / (2x)2
0.016 (4x2) = 0.48 - 1.4x + x2
0.064x2 = 0.48 - 1.4x + x2
0.936x2 -1.4x + 0.48 = 0
x = 0.53 M
Thus, at equilibrium
[H2] = 0.8- 0.53 = 0.27 M
[I2] = 0.6 - 0.53 = 0.07 M
[HI] = 2*0.53 = 1.06 M
4.00 mols of H2 and 3 mols of I2 are placed in an evacuated 5.00L flask...
Suppose that 0.1000 mole each of H2 and I2 are placed in 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equilibrium, the concentration of I2 is found to be 0.0210 M. a) What are the equilibrium concentrations of H2 and HI, respectively? Calculate Kc for the following reaction at 425oC. H2(g) + I2(g) ⇄ 2 HI(g) b) If the initial concentrations of H2 and I2 are 1.000 M each, and the initial concentration of HI is 0.000,...
A flask containing 0.10 atm of H2(g) and excess I2(s) is heated and allowed to reach equilibrium according to the reaction below. The equilibrium constant is 0.25 at this temperature. What is the partial pressure of HI(g) at equilibrium? 2HI(g) --> H2(g) + I2(s) Please help and show work. Thank you
a) If 4.0 moles of H2 and 4.0 moles of I2 are placed in a 4.0 L vessel, how many moles of HI will be present at equilibrium? Final concentration of I2? H2(g) + I2(g) ⇌ 2HI(g) KC = 55.64 @ 425ºC
In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.23 L flask with 0.742 mol of HI gas and allows the reaction to proceed at 426°C: 2HI (g) ⇋ H2(g) + I2(g) At equilibrium, the concentration of HI = 0.084 M. Calculate Kc. Enter to 4 decimal places.
If a 2L reaction vessel initially contains 2 mols of Hydrogen and 2 mols of Iodine, what will be the molar concentrations at equilibrium of all species? H2 (g) + I2 (g) --> 2HI (g) Kc = 50.5
The reaction H2(g) + I2(g) <=> 2HI(g) has Kc = 50.2 at 718K. If a flask is charged with 4.25M HI, what are the equilibrium concentrations of H2 and I2? at equilibrium, [H^2}+[I^2]= ?????
a) In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.79 L flask with 0.452 mol of HI gas and allows the reaction to proceed at 428°C: 2HI (g) ⇋ H2(g) + I2(g) At equilibrium, the concentration of HI = 0.055 M. Calculate Kc. Enter to 4 decimal places. HINT: Look at sample problem 17.6 in the 8th ed Silberberg book. Write a Kc expression. Find the initial concentration. Fill in the ICE chart. Put the E (equilibrium) values...
1. In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.79 L flask with 0.222 mol of HI gas and allows the reaction to proceed at 436°C: 2HI (g) ⇋ H2(g) + I2(g) At equilibrium, the concentration of HI = 0.09 M. Calculate Kc. Enter to 4 decimal places. HINT: Look at sample problem 17.6 in the 8th ed Silberberg book. Write a Kc expression. Find the initial concentration. Fill in the ICE chart. Put the E (equilibrium) values...
H2(g)+I2(g)⇌2HI(g) For the above reaction, Kc=55.3 at 700 K. In a 2.00-L flask containing an equilibrium mixture of the three gases, there are 0.053 g of H2 and 4.39 g of I2. What is the mass of HI in the flask?
For the reaction H2 (g) + I2 (g) = 2HI (g); Kc =50.0. Calculate the concentration of HI (g) at equilibrium if the initial concentration of each substance is 0.0600 M and the reaction mixture is allowed to come to equilibrium. (Hint: ICE Table)