Sodium metal reacts with water to produce hydrogen gas and sodium hydroxide according to the chemical equation shown below.
when 0.0 25 mol of na is added to 100.00 g of water, the temperature of the resulting solution rises from 25.00°c to 35.75°c.

if the specific heat of the solution is 4.18 j/(g · °c), calculate δh for the reaction, as written.

2 na(s) + 2 h2o(l) → 2 naoh(aq) + h2(g) δh= ?

The enthalpy change of the reaction is -361.6 kJ

Explanation:

To calculate the number of moles, we use the equation:

Moles of sodium = 0.025 moles

Molar mass of sodium = 23 g/mol

Putting values in above equation, we get:

We are given:

Mass of water = 100.00 g

Mass of sodium = 0.575 g

Mass of solution = 100.00 + 0.575 = 100.575 g

To calculate the amount of heat absorbed, we use the equation:

where,

q = amount of heat absorbed = ?

m = mass of solution = 100.575 g

C = specific heat capacity of solution = 4.18 J/g°C

= change in temperature =

Putting all the values in above equation, we get:

When heat is absorbed by the solution, this means that heat is getting released by the reaction.

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

When 0.025 moles of sodium is reacted, the heat released by the reaction is 4.52 kJ

So, when 2 moles of sodium will react, the heat released by the reaction will be =

Hence, the enthalpy change of the reaction is -361.6 kJ