Explanation:
A chemical reaction in which heat or energy is released is known as an exothermic reaction.
On the other hand, when two objects are placed together and heat flows from hotter object to colder object then this process is known as conduction. Therefore, energy is dissipated in conduction process.
Since energy released released goes into the atmosphere and is not used anywhere.
Thus, we can conclude that when an exothermic reaction releases thermal energy, this energy is usually not useable to do work and it is dissipated by conduction.
<span>It can form four covalent bonds. </span>
Metamorphic rocks form when pre-existing or parent rocks (whether igneous, sedimentary, or even metamorphic) are altered by heat, pressure, and the chemical activity of fluids. When the dominant altering factor is heat, usually from direct contact , it may undergo fundamental change in texture and recrystallization is called contact metamorphism. While regional metamorphism, where rocks texture and minerals are changed by heat and pressure.
Explanation:
1) N₂ + O₂ → 2 NO
Kc = [NO]² / ([N₂] [O₂])
Set up an ICE table:
![\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\N_{2}&0.114&-x&0.114-x\\O_{2}&0.114&-x&0.114-x\\NO&0&+2x&2x\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26Initial%26Change%26Equilibrium%5C%5CN_%7B2%7D%260.114%26-x%260.114-x%5C%5CO_%7B2%7D%260.114%26-x%260.114-x%5C%5CNO%260%26%2B2x%262x%5Cend%7Barray%7D%5Cright%5D)
Plug into the equilibrium equation and solve for x.
1.00×10⁻⁵ = (2x)² / ((0.114 − x) (0.114 − x))
1.00×10⁻⁵ = (2x)² / (0.114 − x)²
√(1.00×10⁻⁵) = 2x / (0.114 − x)
0.00316 = 2x / (0.114 − x)
0.00361 − 0.00316x = 2x
0.00361 = 2.00316x
x = 0.00018
The volume is 1.00 L, so the concentrations at equilibrium are:
[N₂] = 0.114 − x = 0.11382
[O₂] = 0.114 − x = 0.11382
[NO] = 2x = 0.00036
2(a) Cl₂ → 2 Cl
Kc = [Cl]² / [Cl₂]
![\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\Cl_{2}&2.0&-x&2.0-x\\Cl&0&+2x&2x\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26Initial%26Change%26Equilibrium%5C%5CCl_%7B2%7D%262.0%26-x%262.0-x%5C%5CCl%260%26%2B2x%262x%5Cend%7Barray%7D%5Cright%5D)
1.2×10⁻⁷ = (2x)² / (2 − x)
1.2×10⁻⁷ (2 − x) = 4x²
2.4×10⁻⁷ − 1.2×10⁻⁷ x = 4x²
2.4×10⁻⁷ ≈ 4x²
x² ≈ 6×10⁻⁸
x ≈ 0.000245
2x ≈ 0.00049
2(b) F₂ → 2 F
Kc = [F]² / [F₂]
![\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\F_{2}&2.0&-x&2.0-x\\F&0&+2x&2x\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26Initial%26Change%26Equilibrium%5C%5CF_%7B2%7D%262.0%26-x%262.0-x%5C%5CF%260%26%2B2x%262x%5Cend%7Barray%7D%5Cright%5D)
1.2×10⁻⁴ = (2x)² / (2 − x)
1.2×10⁻⁴ (2 − x) = 4x²
2.4×10⁻⁴ − 1.2×10⁻⁴ x = 4x²
2.4×10⁻⁴ ≈ 4x²
x² ≈ 6×10⁻⁵
x ≈ 0.00775
2x ≈ 0.0155
F₂ dissociates more, so Cl₂ is more stable at 1000 K.
Answer:
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