The reaction that occurs between a strong monoprotic acid and sodium hydroxide is H++OH- => H2O. This reaction is the same for all strong monoprotic acids and sodium hydroxide so, in theory, they should all have the same standard enthalpy of reaction. In practice, there are very slight differences between acids. If you are in a freshman or sophmore chemistry class, say yes. If you are in physical or analytical chemistry say no.
The enthalpy of a chemical reaction is the change of heat during this reaction.
The enthalpy of a chemical reaction is the change of heat during this reaction.
No, a catalyst will not change reaction enthalpy. If it does so, then it is NOT a catalyst but a reactant in stead!
Absolutely not.
True, a large positive value of entropy tends to favor products of a chemical reaction. However, entropy can be offset by enthalpy; a large positive value of enthalpy tends to favor the reactants of a chemical reaction. The true measure to determine which side of a chemical reaction is favored is the change in Gibbs' free energy, which accounts for both entropy and enthalpy, as calculated by: Change in Gibbs = Change in Enthalpy - Temp in Kelvin * Change in Entropy A negative value of Gibbs free energy will always favour the products of a chemical reaction.
The enthalpy of reaction is the change of the system enthalpy after a chemical reaction.
The enthalpy of a chemical reaction is the change of heat during this reaction.
The enthalpy of a chemical reaction is the change of heat during this reaction.
The enthalpy of a chemical reaction is the change of heat during this reaction.
The enthalpy of a chemical reaction is the change of heat during this reaction.
No, a catalyst will not change reaction enthalpy. If it does so, then it is NOT a catalyst but a reactant in stead!
The enthalpy of a reaction does not depend on the reactant path taken.
Absolutely not.
An exothermic reaction is a chemical reaction that releases energy in the form of heat. It favors a negative enthalpy change.
All the reactions in a path are added together.
Yes, the reaction involving the solid is actually an individual step in the equation of the reaction between the solutions. If you were to add the change in enthalpy of the reaction with the solid NaOh to the change in enthalpy of the other step in the reaction (that's adding water and the NaOh pellets) you would find the sum equivalent to the change in enthalpy of the reaction involving the two solutions (this is supported by Hess's law). I suggest that you consider Hess's law for more information.
The amount of energy that is lost or gained by the products during the reaction.