Hydrogen reacts with iodine to form hydrogen iodide as follows:

H2 (g) + I2 (g) ⇌ 2 HI (g)

At 445 °C, Kc = 50.2. If one starts with 0.100 M H2 (g), 0.100 M I2 (g) and 0.0500 M HI (g) what is the equilibrium concentration of HI?

Question

Hydrogen reacts with iodine to form hydrogen iodide as follows:

H2 (g) + I2 (g)  ⇌  2 HI (g)

At 445 °C, Kc = 50.2. If one starts with 0.100 M H2 (g), 0.100 M I2 (g) and 0.0500 M HI (g) what is the equilibrium concentration of HI?

lena772 · Answer

@Photon336

photon336 · Answer

$$Kc = \frac{ [HI]^{2} }{ [H_{2}][I_{2}] }$$

photon336 · Answer

I would also calculate K to be sure of what direction our reaction is in. 

$$\frac{ [0.05]^{2} }{ [0.10][0.10] }  = 0.25 Q $$

Q < K so the reaction is going to go to the right

photon336 · Answer

I guess we could do an ice table for this

lena772 · Answer

0.25

photon336 · Answer

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