Question

Hydrogen iodide decomposes according to the following equation:
H2(g) + I2(g) ⇌ 2HI(g)
A 2.50 L equilibrium mixture contains 0.0875 mol H2, 0.020 mol I2, and 3.72 mol HI. Determine the new equilibrium concentrations after 1.25 mol HI are added to the system.


This is what I've done so far, I made this ICE table:

IE (M): 0.0350 0.0800 1.488
D(M): +0.500
NI(M): 0.0350 0.0800 1.988
C(M): -x -x +2x
NE(M): 0.0350-x 0.0800-x 1.988+2x

Is this correct? what do I do next?

Answer 1

First, that 0.020/2.5L = 0.008 and not 0.08.
Second, you wrote the equation as a synthesis and not a decomposition. That doesn't make that much difference as long as you keep Keq straight.
Third, if you add 0.5 M HI to the system which is already at equilibrium, the system will shift away from HI and to H2 and I2; therefore, those should be +x and +x and HI is not -2x/
Fourth, you should calculate Keq for the system as you view it; i.e., as a decomposition or as a synthesis.
Post your work if you get stuck.