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...
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