Answer:
b. Second order in NO and first order in O₂.
Explanation:
A. The mechanism
![\rm 2NO\xrightarrow[k_{-1}]{k_{1}}N_{2}O_{2} \, (fast)\\\rm N_{2}O_{2} + O_{2}\xrightarrow{k_{2}} 2NO_{2} \, (slow)](https://tex.z-dn.net/?f=%5Crm%202NO%5Cxrightarrow%5Bk_%7B-1%7D%5D%7Bk_%7B1%7D%7DN_%7B2%7DO_%7B2%7D%20%5C%2C%20%28fast%29%5C%5C%5Crm%20N_%7B2%7DO_%7B2%7D%20%2B%20O_%7B2%7D%5Cxrightarrow%7Bk_%7B2%7D%7D%202NO_%7B2%7D%20%5C%2C%20%28slow%29)
B. The rate expressions
![-\dfrac{\text{d[NO]} }{\text{d}t} = k_{1}[\text{NO]}^{2} - k_{-1} [\text{N}_{2}\text{O}_{2}]^{2}\\\\\rm -\dfrac{\text{d[N$_{2}$O$_{2}$]}}{\text{d}t} = -\dfrac{\text{d[O$_{2}$]}}{\text{d}t} = k_{2}[ N_{2}O_{2}][O_{2}] - k_{1} [NO]^{2}\\\\\dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= k_{2}[ N_{2}O_{2}][O_{2}]](https://tex.z-dn.net/?f=-%5Cdfrac%7B%5Ctext%7Bd%5BNO%5D%7D%20%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B1%7D%5B%5Ctext%7BNO%5D%7D%5E%7B2%7D%20-%20k_%7B-1%7D%20%5B%5Ctext%7BN%7D_%7B2%7D%5Ctext%7BO%7D_%7B2%7D%5D%5E%7B2%7D%5C%5C%5C%5C%5Crm%20-%5Cdfrac%7B%5Ctext%7Bd%5BN%24_%7B2%7D%24O%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20-%5Cdfrac%7B%5Ctext%7Bd%5BO%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B2%7D%5B%20N_%7B2%7DO_%7B2%7D%5D%5BO_%7B2%7D%5D%20-%20k_%7B1%7D%20%5BNO%5D%5E%7B2%7D%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7Bd%5BNO%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%3D%20k_%7B2%7D%5B%20N_%7B2%7DO_%7B2%7D%5D%5BO_%7B2%7D%5D)
The last expression is the rate law for the slow step. However, it contains the intermediate N₂O₂, so it can't be the final answer.
C. Assume the first step is an equilibrium
If the first step is an equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium is only slightly perturbed by the slow leaking away of N₂O₂ to form product.
![\rm k_{1}[NO]^{2} = k_{-1} [N_{2}O_{2}]\\\\\rm [N_{2}O_{2}] = \dfrac{k_{1}}{k_{-1}}[NO]^{2}](https://tex.z-dn.net/?f=%5Crm%20k_%7B1%7D%5BNO%5D%5E%7B2%7D%20%3D%20k_%7B-1%7D%20%5BN_%7B2%7DO_%7B2%7D%5D%5C%5C%5C%5C%5Crm%20%5BN_%7B2%7DO_%7B2%7D%5D%20%3D%20%5Cdfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5E%7B2%7D)
D. Substitute this concentration into the rate law
![\rm \dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= \dfrac{k_{2}k_{1}}{k_{-1}}[NO]^{2} [O_{2}] = k[NO]^{2} [O_{2}]](https://tex.z-dn.net/?f=%5Crm%20%5Cdfrac%7B%5Ctext%7Bd%5BNO%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%3D%20%5Cdfrac%7Bk_%7B2%7Dk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5E%7B2%7D%20%5BO_%7B2%7D%5D%20%3D%20k%5BNO%5D%5E%7B2%7D%20%5BO_%7B2%7D%5D)
The reaction is second order in NO and first order in O₂.
Explanation:
so, as you move down a group on the periodic table, the electronegativity of an element decreases because the increased number of energy levels puts the outer electrons very far away from the pull of the nucleus. electronegativity increases as you move from left to right across a period on the periodic table.
Here we are talking about ammonia which is a molecule. Now 1 mole of ammonia also contains 6.022<span> * 10^</span>23 molecules<span> of ammonia. In one molecule of ammonia,we have 1 nitrogen atom and 3 hydrogen atom.</span>
36.5ml is the new concentration so, its a chemical change if there's something new
