Question

When 0.717 g of sodium metal is added to an excess of hydrochloric acid, 7450 J of heat are produced. What is the enthalpy of the reaction as written?

2Na(s)+2HCl(aq)⟶2NaCl(aq)+H2(g)

Answer 1

Answer:

The enthalpy change of the reaction is 4.78 × 10⁴ J.

Explanation:

We are given that 0.717 g of sodium metal reacts with hydrochloric acid to produce 7450 J of heat.

Converting grams of sodium to moles:

$\displaystyle 0.717 \text{ g Na} \cdot \frac{1 \text{ mol Na}}{22.99 \text{ g Na}} = 0.0312 \text{ mol Na}$

And dividing the amount of heat produced by the moles of sodium reacted yields:

$\displaystyle \Delta H = \frac{7450 \text{ J}}{0.312 \text{ mol Na}} = 2.39 \times 10^4 \text{ J/mol Na}$

Because the given reaction has two moles of sodium metal, we can multiply the above value by two to acquire the enthalpy change of the given reaction:

$\displaystyle \Delta H = 2\text{ mol Na}\left(2.39 \times 10^4 \text{ J/mol Na}}\right) = 4.78 \times 10^4 \text{ J}$

In conclusion, the enthalpy change of the reaction is 4.78 × 10⁴ J.