Question

At 500 °C, hydrogen iodide decomposes according to 2HI(g) ↽−−⇀ H₂(g) + I₂(g) For HI(g) heated to 500 °C in a 1.00 L reaction vessel, chemical analysis determined these concentrations at equilibrium: [H₂]=0.307 M , [I₂]=0.307 M , and [HI]=2.59 M . If an additional 1.00 mol of HI(g) is introduced into the reaction vessel, what are the equilibrium concentrations after the new equilibrium has been reached?

Answer 1

Answer:

$[HI]_{eq}=2.902M$

$[I_2]_{eq}=0.344M$

$[H_2]_{eq}=0.344M$

Explanation:

Hello,

At first, with the equilibrium concentrations we compute the equilibrium constant as:

$K=\frac{[H_2]_{eq}[I_2]_{eq}}{[HI]_{eq}^2}=\frac{(0.307M)(0.307M)}{(2.59M)^2} =0.0141$

Now, as 1 mol of HI is added, it has now a 3.59-M concentration, so one modifies the equilibrium based on the new conditions and change, x:

$0.0141=\frac{(x)(x)}{(3.59M-2x)^2} \\0.0141=\frac{x^2}{12.89-3.59x+4x^2} \\0.0141(12.89-14.36x+4x^2)=x^2\\0.1817-0.2025x-0.9436x^2=0\\x_1=-0.559M\\x_2=0.344M$

Finally, the new equilibrium concentrations are:

$[HI]_{eq}=3.59M-2(0.344M)=2.902M$

$[I_2]_{eq}=0.344M$

$[H_2]_{eq}=0.344M$

Best regards.