The organic compound di-n-butyl phthalate, C16H22O4(l), is sometimes used as a low-density (1.046 g·mL–1) manometer fluid. Compu
2 answers:
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Answer:
124.91mL
Explanation:
Given parameters:
P₁ = 1.08atm
V₁ = 250mL
T₁ = 24°C
P₂ = 2.25atm
T₂ = 37.2°C
V₂ = ?
Solution:
To solve this problem, we are going to apply the combined gas law;
P, V and T represents pressure, volume and temperature
1 and 2 delineates initial and final states
Convert the temperature to kelvin;
T₁ = 24°C, T₁ = 24 + 273 = 297K
T₂ = 37.2°C , T₂ = 37.2 + 273 = 310.2K
Input the variables and solve for V₂
V₂ = 124.91mL
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 correct option is, (a) 0.44
Explanation :
First we have to calculate the concentration of .
Now we have to calculate the dissociated concentration of .
The balanced equilibrium reaction is,
Initial conc. 1.0 M 0
At eqm. conc. (1.0-x) M (2x) M
As we are given,
The percent of dissociation of =
= 28.0 %
So, the dissociate concentration of =
The value of x = = 0.28 M
Now we have to calculate the concentration of at equilibrium.
Concentration of = 1.0 - x = 1.0 - 0.28 = 0.72 M
Concentration of = 2x = 2 × 0.28 = 0.56 M
Now we have to calculate the equilibrium constant for the reaction.
The expression of equilibrium constant for the reaction will be:
Now put all the values in this expression, we get :
Therefore, the equilibrium constant for the reaction is, 0.44