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Why is useful to use moles to measure chemical quantities
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>
Answer:
The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³
(Option C)
Explanation:
Given;
concentration of HA, = 6.0mol/dm³
volume of HA, = 25.0cm³, = 0.025dm³
Concentration of HB, = 3.0mol/dm³
volume of HB, = 45.0cm³ = 0.045dm³
To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;
where;
is initial concentration
is initial volume
is final concentration of the solution
is final volume of the solution
Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³