Question

Which of the following has the correct change in enthalpy and reaction description for the following reaction?

Answer 1

delta is always the difference between the final minus the initial. In this case the delta H of the reaction you have to add the delta H of formation of the products and subtract from them the sum of the delta H of formation of the reactants, then if the final answer is a negative number that means that the energy of the products is lower than the energy of the reactants then you can figure out than the reaction lost energy. All the options are negative. So you can discard without doing too much calculations a few answers.

Answer 2

Could you like explain to me which one would be formation of products?

Answer 3

in any chemical reaction you have always at your left the reactants then an arrow and at the right of the arrow the products of the reaction.
In your case we can said the NH3 and the HCl are the reactants of that reaction and the NH4Cl is the product of your reaction.
Then you have a series of delta H given for each of your compounds. They are the delta H of formation of each of them (I think so).
The represent the energy used to produce one mole of that compound. In the case of your reaction all the compounds stoichiometry is one to one, if you have for example 2 moles of HCl you will have to multiply that delta H by 2 before add to the delta H of the NH3 (but this is not your case now)

Answer 4

This is similar to your problem

Calculate deltaH for the following reaction:

NH4NO3(s) + H2O(l) ---> NH4+ (aq) + NO3- (aq)

Use the results of this calculation to determine the value of deltaGo for this reaction at 25o C, and explain why NH4NO3 spontaneously dissolves is water at room temperature.

Solution

Using a standard-state enthalpy of formation and absolute entropy data table, we find the following information:

Compound deltaHfo(kJ/mol)

NH4NO3(s) -365.56

NH4+ (aq) -132.51

NO3- (aq) -205.0

This reaction is endothermic, and the enthalpy of reaction is therefore unfavorable:

deltaHo = sigmaHfo(products) - sigmaHfo(reactants)

= [1 mol NH4 x 132.51 kJ/mol + 1 mol NO3- x -205.0 kJ/mol] - [1 mol NH4NO3 x -365.56 kJ/mol]

= 28.05 kJ

Answer 5

Okay I came up with option A

Answer 6

I think that too!