You place 40.6 ml of 0.659 m naoh in a coffee- cup calorimeter at 25.00°c and add 71.3 ml of 0.659 m hcl, also at 25.00°c. after stirring, the final temperature is 28.57°c. [assume the total volume is the sum of the individual volumes and that the final solution has the same density (1.00 g/ml) and specific heat capacity (4.184 j/gk)]. calculate the change in enthalpy (latex: \deltaδh) of the reaction in kj/mol of water formed. enter to 1 decimal place.

ΔH of reaction is 62,3 kJ/mol

Explanation:

For the reaction:

HCl + NaOH → H₂O + NaCl

The heat transfer in the reaction (q) =

q = Specific heat × mass × ΔT

The mass is 40,6 mL + 71,3 mL × (1g/mL) = 111,9 g

And ΔT is 28,57 - 25,00 = 3,57°C

q = 4,184 J/gK × 111,9 g × 3,57°C = 1671 J

The moles of NaOH are:

0,0406 L×0,659 M = 0,0268 moles

And the moles of HCl are:

0,0713 L ×0,659 M = 0,0470 moles

The moles of reaction are the moles of NaOH because are the limiting reactant. 0,0268 moles

ΔH of reaction is q/moles of reaction. Thus:

ΔH of reaction =  1671J/0,0268 moles = 62350 J/mol ≡ 62,3 kJ/mol

I hope it helps!

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

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