Answer: Rate law=![k[A]^1[B]^2](https://tex.z-dn.net/?f=k%5BA%5D%5E1%5BB%5D%5E2)
, order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is 
Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
![Rate=k[A]^x[B]^y](https://tex.z-dn.net/?f=Rate%3Dk%5BA%5D%5Ex%5BB%5D%5Ey)
k= rate constant
x = order with respect to A
y = order with respect to A
n = x+y = Total order
a) From trial 1: ![1.2\times 10^{-2}=k[0.10]^x[0.20]^y](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E%7B-2%7D%3Dk%5B0.10%5D%5Ex%5B0.20%5D%5Ey)
(1)
From trial 2: ![4.8\times 10^{-2}=k[0.10]^x[0.40]^y](https://tex.z-dn.net/?f=4.8%5Ctimes%2010%5E%7B-2%7D%3Dk%5B0.10%5D%5Ex%5B0.40%5D%5Ey)
(2)
Dividing 2 by 1 :![\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}](https://tex.z-dn.net/?f=%5Cfrac%7B4.8%5Ctimes%2010%5E%7B-2%7D%7D%7B1.2%5Ctimes%2010%5E%7B-2%7D%7D%3D%5Cfrac%7Bk%5B0.10%5D%5Ex%5B0.40%5D%5Ey%7D%7Bk%5B0.10%5D%5Ex%5B0.20%5D%5Ey%7D)
therefore y=2.
b) From trial 2: (3)
From trial 3: ![9.6\times 10^{-2}=k[0.20]^x[0.40]^y](https://tex.z-dn.net/?f=9.6%5Ctimes%2010%5E%7B-2%7D%3Dk%5B0.20%5D%5Ex%5B0.40%5D%5Ey)
(4)
Dividing 4 by 3:![\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}](https://tex.z-dn.net/?f=%5Cfrac%7B9.6%5Ctimes%2010%5E%7B-2%7D%7D%7B4.8%5Ctimes%2010%5E%7B-2%7D%7D%3D%5Cfrac%7Bk%5B0.20%5D%5Ex%5B0.40%5D%5Ey%7D%7Bk%5B0.10%5D%5Ex%5B0.40%5D%5Ey%7D)
, x=1
Thus rate law is ![Rate=k[A]^1[B]^2](https://tex.z-dn.net/?f=Rate%3Dk%5BA%5D%5E1%5BB%5D%5E2)
Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.
c) For calculating k:
Using trial 1: ![1.2\times 10^{-2}=k[0.10]^1[0.20]^2](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E%7B-2%7D%3Dk%5B0.10%5D%5E1%5B0.20%5D%5E2)
.
Answer:
Explanation:
The period law state that when elements are listed in order of their atomic numbers, the elements fall into recurring groups, so that there is a recurrence of similar properties at regular intervals.
Na and K in the periodic table fall into the same group, this is because they both have one electrons in their outermost shell.
Na 11 -1s2 2s2 2p6 3s1
K 19 - 1s2 2s2 2p6 3s2 3p6 4s1
They share similar chemical and physical properties. Na and K are very reactive metals, they can loose/donate their outermost electron to non metals in other to attain stable octet state.
The form ionic compound when they react with non metals.
Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
