Find the enthalpy of neutralization of hcl and naoh. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6 mol dm-3 naoh. the temperature rose from 298 k to 325.8 k. the specific heat capacity is the same as water, 4.18 j/k g.
a. 44.69 kj/mol
b. 6123.06 kj/mol
c. 597.46 kj/mol
d. 89.39 kj/mol

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

The balanced chemical reaction will be,

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water =

Now we have to calculate the heat absorbed during the reaction.

where,

q = heat absorbed = ?

= specific heat of water =

m = mass of water = 274 g

= final temperature of water = 325.8 K

= initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

where,

= enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).