How are trials used in an experiment?
1 answer:
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Answer 1:
Equilibrium constant (K) mathematically expressed as the ratio of the concentration of products to concentration of reactant. In case of gaseous system, partial pressure is used, instead to concentration.
In present case, following reaction is involved:
2NO2 ↔ 2NO + O2
Here, K =![\frac{[PNO]^2[O2]}{[PNO2]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BPNO%5D%5E2%5BO2%5D%7D%7B%5BPNO2%5D%5E2%7D%20)
Given: At equilibrium, <span>PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm
</span>
Hence, K =![\frac{[0.0022]^2[0.0011]}{[0.247]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5B0.0022%5D%5E2%5B0.0011%5D%7D%7B%5B0.247%5D%5E2%7D%20)
= 8.727 X 10^-8
Thus, equilibrium constant of reaction = 8.727 X 10^-8
.......................................................................................................................
Answer 2:
Given: <span>PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm.
Therefore, Reaction quotient = </span>
=![\frac{[0.021]^2[0.037]}{[0.192]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5B0.021%5D%5E2%5B0.037%5D%7D%7B%5B0.192%5D%5E2%7D%20)
= 4.426 X 10^-4.
Here, Reaction quotient > Equilibrium constant.
Hence, <span>the reaction need to go to reverse direction to reattain equilibrium </span>
Equilibrium constant (K) mathematically expressed as the ratio of the concentration of products to concentration of reactant. In case of gaseous system, partial pressure is used, instead to concentration.
In present case, following reaction is involved:
2NO2 ↔ 2NO + O2
Here, K =
Given: At equilibrium, <span>PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm
</span>
Hence, K =
= 8.727 X 10^-8
Thus, equilibrium constant of reaction = 8.727 X 10^-8
.......................................................................................................................
Answer 2:
Given: <span>PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm.
Therefore, Reaction quotient = </span>
=
= 4.426 X 10^-4.
Here, Reaction quotient > Equilibrium constant.
Hence, <span>the reaction need to go to reverse direction to reattain equilibrium </span>