What element is in group 17 and period 3 of the periodic table?
2 answers:
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Answer:
a)
b)
Explanation:
Equation of reaction:
Initial pressure 3 1 0
Pressure change 2P 1P 2P
Total pressure = (3-2P) + (1-P) + (2P)
Total Pressure = 3.75 atm
(3-2P) + (1-P) + (2P) = 3.75
4 - P = 3.75
P = 4 - 3.75
P = 0.25 atm
Let us calculate the pressure of each of the components of the reaction:
Pressure of XO2 = 3 - 2P = 3 - 2(0.25)
Pressure of XO2 =2.5 atm
Pressure of O2 = 1 - P = 1 -0.25
Pressure of O2 = 0.75 atm
Pressure of XO3 = 2P = 2 * 0.25
Pressure of XO3 = 0.5 atm
From the reaction, equilibrium constant can be calculated using the formula:
Standard free energy:
b) value of k−1 at 27 °C, i.e. 300K
Answer:
concentration of = 0.0124 = 12.4 ×10⁻³ M
concentration of = 0.0248 = 2.48 ×10⁻² M
concentration of = 0.4442 M
Explanation:
Equation for the reaction:
⇄
+
Concentration of =
= 0.469
For our ICE Table; we have:
⇄
+
Initial 0.469 0 0
Change - 2x +2x +x
Equilibrium (0.469-2x) 2x x
K =
K =
Since the value pf K is very small, only little small of reactant goes into product; so (0.469-2x)² = (0.469)²
x = 0.0124
∴ at equilibrium; concentration of = 0.0124 = 12.4 ×10⁻³ M
concentration of = 2x = 2 ( 0.0124)
= 0.0248
= 2.48 ×10⁻² M
concentration of = 0.469-2x
= 0.469-2(0.0124)
= 0.469 - 0.0248
= 0.4442 M