Answer:
-0.85KJ
Explanation:
Given N2(g) + H2(g) <--->2NH3(g)
Kp =[ P(NH3)]²/[P(H2)]³[P(N2)]
Where P is the pressure of the gas
P(H2)b= P(N2) = 125atm
P(NH3) = 200atm
Kp = 2²/(125)³(125)
Kp = 2.048 ×10^-6
∆G = -RTlnKp
R =0.008314 J/Kmol
T = 25 +273/= 298k
= 8.314 ×10^-3 × 298 × ln(2.048 ×10^-6)
= -0.008314 × 298 × (-13.099)
= 32.45KJ
∆G = ∆G° + RTlnKp
∆G = -33.3 + 32.45
∆G = -0.85KJ or -850J
Answer: The pH at the equivalence point for the titration will be 0.65.
Solution:
Let the concentration of
be x
Initial concentration of
, c = 0.230 M

at eq'm c-x x x
Expression of
:
![K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5B%2BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D%3D%5Cfrac%7Bx%5Ctimes%20x%7D%7Bc-x%7D%3D%5Cfrac%7Bx%5E2%7D%7Bc-x%7D)
Since ,methyl-amine is a weak base,c>>x so
.

Solving for x, we get:

Given, HCl with 0.230 M , it dissociates fully in water which means
= 0.230 M
will result in neutral solution, since ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3C%5BH%5E%2B%5D)
Remaining
after neutralizing
ions
![[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D%5BH%5E%2B%5D-%5BOH%5E-%5D%3D0.230-1.07%5Ctimes%2010%5E%7B-2%7D%3D0.2193%20M)
![pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65](https://tex.z-dn.net/?f=pH%3D-log%7B%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D-log%280.2193%29%3D0.65)
The pH at the equivalence point for the titration will be 0.65.
Answer:
23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr
Explanation:
The balanced equation here is
6NaBr + 1AlO3 = 3Na2O + 2AlBr3
6 moles of NaBr are required to produce 2 moles of AlBr3
Mass of one mole of NaBr = 102.894 g/mol
Mass of one mole of AlBr3 = 266.69 g/mol
Mass of 6 moles of NaBr = 6*102.894 g/mol
Mass of two moles of AlBr3 = 2*266.69 g/mol
6*102.894 g NaBr produces 2*266.69 g of AlBr3
23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr