# Question #eb3dd

May 1, 2017

206.00 $k J m o {l}^{- 1}$

#### Explanation:

The formula to calculate the heat change of a reaction is
ΔH = -mcΔT (where m is the mass of the fluid whose temperature changes which in this case is water, c is the specific heat capacity of water and ΔT is the change in temperature of water)

Heat change when 0.025 mol of Na is added
= $- 100 \cdot 4.18 \cdot \left(35.75 - 25.00\right)$
= $- 5138.5 J$

Now this is heat change for 0.025 mole of Na.
Enthalpy change means you have to calculate the heat change for 1 mole.

So enthalpy change for 1 mole of Na is $- \frac{5138.5}{0.025}$ = 205540 J= 206.00 $k J m o {l}^{- 1}$