Colder- moves slower
warmer- moves faster
changes state:
solid to gas- warmer
gas to solid- colder
something like that
Question:
<em>What effects does the concentration of reactants have on the rate of a reaction?</em>
Answer:
<em>Reactant concentration. Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.</em>
<em>Increasing the concentration of reactants generally increases the rate of reaction because more of the reacting molecules or ions are present to form the reaction products. ... When concentrations are already high, a limit is often reached where increasing the concentration has little effect on the rate of reaction.</em>
Hope this helps, have a good day. c;
Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
<span>the answer is organic molecule </span>
