What class is this? (Subject)
Wave length measures the width of the wave
Answer:
— The molality of chloride ions in 300g of water is. A) 1.00 molal. B) 0.500 molal. C) 0.0553 molal. D) 0.111 molal.
Answer:
The molecules absorb heat and acquire more kinetic energy.
Explanation:
In a solid, the solids only vibrate about their mean positions but do not translate. When energy is supplied to the molecule in the form of heat, the molecules vibrate faster. Eventually, they acquire sufficient energy to leave their mean positions and translate. Hence the solid crystal collapses.
When ice is heated, water molecules acquire sufficient kinetic energy to translate. The intermolecular bonds are gradually broken in the solid framework as heat is absorbed. The heat required for this is known as the latent heat of fusion.
The temperature remains constant until phase transition is over, then temperature rise resumes.
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.