₁₇Cl 1s²2s²2p⁶3s²3p⁵, 1 unpaired electron in 3pz orbital.
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:
![K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BPXO_%7B3%7D%5D%20%5E%7B2%7D%20%7D%7B%5BPXO_%7B2%7D%5D%20%5E%7B2%7D%5BPO_%7B2%7D%5D%20%7D)
Standard free energy:
b) value of k−1 at 27 °C, i.e. 300K
Answer:
Explanation:
mole of NaOH present = molarity x volume
= 1.0 X 0.05 = 0.05 mole
<em>Recommended mole of HCl </em>= 1.1 x 0.05 = 0.055
<em>Mole of HCl carelessly added by Jacob </em>= 1.1 x 0.04 = 0.044
From the equation of reaction:
HCl + NaOH ----> NaCl + H2O
The ratio of mole of HCl to that of NaOH for a complete neutralization reaction is 1:1. However, the recommended mole of HCl (0.055 mole) is more than the mole of NaOH (0.05 mole). <u>Hence, the recommended endpoint of the reaction is supposed to be acidic.</u>
The mole of HCl added by Jacob (0.044) is short of the recommended amount (0.055) and also short of the amount required for a neutral endpoint (0.05). <u>This means that the endpoint will have an excess amount of NaOH and as such, basic instead of the desired acidic endpoint.</u>
