When 50.0 ml of 0.500 m hcl at 25.0°c is added to 50.0 ml of 0.500 m naoh at 25.0°c in a coffee cup calorimeter, the temperature of the mixture rises to 28.2°c.

what is the enthalpy of reaction per mole of acid?

assume the mixture has a specific heat capacity of 4.18 j/(g ? k) and that the densities of the reactant solutions are both 1.00 g/ml.

enthalpy of reaction =  -53504.00 J/mol = -53.50 kJ/mol

Explanation:

The heat released in the reaction is the heat absorbed by the calorimeter. The heat absorbed by the calorimeter is given by the expresion:

Qcal = mx c x ΔT where

c = specific heat capacity = 4.18 J/gK

ΔT = the change in temperature =  (28.2 - 25.5 ) K  (in this case we do not need to convert  ºC to K since what we need is the change)

Since the enthalpy of enthalpy of the reaction is the negative of Q we will finally need to convert this heat per mol HCl ( this is because we are told 0.025 mol HCl were reacted)

m = (50.0 + 50.0 mL) x 1g/mL = 100 g

Q = 100 g x 4.18 J/Kmol x 3.2 K = 1337.6 J

V HCl = 50.0 mL /1000 mL/L x 0.5 M = 0.025 mol

enthalpy = - 1337.6 J / 0.025 mol = -53504.00 J/mol = -53.50 kJ/mol

Explanation:

Hello,

In this case, by cause of the temperature change it is possible to compute that reaction's enthalpy of reaction as shown below:

In such a way, by knowing that the mass results from adding both solutions' volumes, we obtain:

Now, the moles of hydrochloric acid are computed as follows:

Finally, the enthalpy of reaction per mole of acid turns out:

Taking into account that it is negative since heat is released due to the temperature increase.

Best regards.

the answer is d. 74.02