Question

At a certain temperature, the equilibrium constant, K c , for this reaction is 53.3. H 2 ( g ) + I 2 ( g ) − ⇀ ↽ − 2 HI ( g ) K c = 53.3 At this temperature, 0.600 mol H 2 and 0.600 mol I 2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?

Answer 1

Answer:

$[HI]_{eq}=0.942M$

Explanation:

Hello,

In this case, the initial concentrations of hydrogen and iodine are the same:

$[H_2]_0=[I_2]_0=0.600M$

Thus, considering the given undergoing chemical reaction, one states the law of mass action in terms of the change $x$ due to the chemical change as shown below:

$Kc=\frac{(2x)^2}{(0.600M-x)(0.600M-x)}=53.3$

Therefore, solving for $x$ by quadratic equation one obtains:

$x_1=0.471M;x_2=0.826M$

Nevertheless, the feasible result is the first one as the second one results in negative concentrations, thus, the hydroiodic acid equilibrium concentration turns out:

$[HI]_{eq}=2*0.471M=0.942M$

Best regards.

Answer 2

Explanation:

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