Question

Sodium metal reacts violently with water in highly exothermic reaction. 2 Na(s) + 2 H2O(l) --> 2 NaOH(aq) + H2(g) ΔHrxn = –368.6 kJ What is the amount of ΔHrxn when 6.00 moles of NaOH is produced from Na and H2O? Group of answer choices

Answer 1

Answer: The value of enthalpy change when 6 moles of NaOH is produced is -1106.4 kJ

Explanation:

We are given:

Moles of NaOH produced = 6.00 moles

For the given chemical equation:

$2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g);\Delta H_{rxn}=-368.6kJ$

The enthalpy change of the reaction = -368.6 kJ

To calculate the amount of enthalpy change, we use unitary method:

When 2 moles of NaOH is produced, the enthalpy change of the reaction is -368.6 kJ

So, when 6 moles of NaOH is produced, the enthalpy change of the reaction will be = $\frac{-368.6}{2}\times 6=-1106.4kJ$

Hence, the value of enthalpy change when 6 moles of NaOH is produced is -1106.4 kJ