This is an incomplete question, here is a complete question.
Hydrogen and iodine react to form hydrogen iodide, like this:
![H_2(g)+I_2(g)\rightarrow 2HI(g)](https://tex.z-dn.net/?f=H_2%28g%29%2BI_2%28g%29%5Crightarrow%202HI%28g%29)
Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition:
Compound Pressure at equilibrium
61.8 atm
46.5 atm
52.3 atm
Calculate the value of the equilibrium constant
for this reaction. Round your answer to 2 significant digits.
Answer : The value of equilibrium constant
for this reaction is, 0.952
Explanation :
The given chemical reaction :
![H_2(g)+I_2(g)\rightarrow 2HI(g)](https://tex.z-dn.net/?f=H_2%28g%29%2BI_2%28g%29%5Crightarrow%202HI%28g%29)
The expression of
for above reaction follows:
![K_p=\frac{(P_{HI})^2}{P_{H_2}\times P_{I_2}}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%28P_%7BHI%7D%29%5E2%7D%7BP_%7BH_2%7D%5Ctimes%20P_%7BI_2%7D%7D)
We are given:
![P_{H_2}=61.8atm](https://tex.z-dn.net/?f=P_%7BH_2%7D%3D61.8atm)
![P_{I_2}=46.5atm](https://tex.z-dn.net/?f=P_%7BI_2%7D%3D46.5atm)
![P_{HI}=52.3atm](https://tex.z-dn.net/?f=P_%7BHI%7D%3D52.3atm)
Putting values in above equation, we get:
![K_p=\frac{(52.3)^2}{61.8\times 46.5}\\\\K_p=0.952](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%2852.3%29%5E2%7D%7B61.8%5Ctimes%2046.5%7D%5C%5C%5C%5CK_p%3D0.952)
Therefore, the value of equilibrium constant
for this reaction is, 0.952