Answer:
Reaction 1: Kc increases
Reaction 2: Kc decreases
Reaction 3: The is no change
Explanation:
Let us consider the following reactions:
Reaction 1: A ⇌ 2B ΔH° = 20.0 kJ/mol
Reaction 2: A + B ⇌ C ΔH° = −5.4 kJ/mol
Reaction 3: 2A⇌ B ΔH° = 0.0 kJ/mol
To predict what will happen when the temperature is raised we need to take into account Le Chatelier Principle: when a system at equilibrium suffers a perturbation, it will shift its equilibrium to counteract such perturbation. This means that <em>if the temperature is raised (perturbation), the system will react to lower the temperature.</em>
Reaction 1 is endothermic (ΔH° > 0). If the temperature is raised the system will favor the forward reaction to absorb heat and lower the temperature, thus increasing the value of Kc.
Reaction 2 is exothermic (ΔH° < 0). If the temperature is raised the system will favor the reverse reaction to absorb heat and lower the temperature, thus decreasing the value of Kc.
Reaction 3 is not endothermic nor exothermic (ΔH° = 0) so an increase in the temperature will have no effect on the equilibrium.
4.) an element cannot be decomposed by a chemical change, it is a pure substance and no other materials compose it.
Answer:
V = 43.95 L
Explanation:
Given data:
Mass of CH₄ decomposed = 15.63 g
Volume of H₂O produced at STP = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → 2H₂O + CO₂
Number of moles of CH₄:
Number of moles = mass/molar mass
Number of moles = 15.63 g/ 16 g/mol
Number of moles = 0.98 mol
Now we will compare the moles of H₂O with CH₄.
CH₄ : H₂O
1 : 2
0.98 : 2×0.98 = 1.96 mol
Volume of hydrogen:
PV = nRT
1 atm × V = 1.96 mol × 0.0821 atm.L/mol.K × 273.15 K
V = 43.95atm.L / 1atm
V = 43.95 L
Answer: a
Explanation:
Industry uses only about 18% while the others use around 70-90% of water.
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