What is the standard enthelpy of formation of NaF

[deuce1906]

Homework Statement

Please help! i just need to see one step by step solution of how to solve this problem and then I can figure the rest out.
When .560g of Na (s) reacts with excess F2(g) to form NaF(s), 13.8 kj of heat is evolved at standard state conditions. What is the standard enthalpy of formation of NaF (s)?

Homework Equations

(delta H=-569Kj/Mol)

The Attempt at a Solution

 

[Borek]

Standard enthalpy is per mole NaF produced, you have it per 0.560g of Na. How many moles produced?

 

[Blueshift5]

To solve this problem, we need to use the formula for standard enthalpy of formation, which is defined as the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states.

The standard state for an element is its most stable form at 1 atm pressure and a specified temperature (usually 25°C). In this case, the standard state for NaF is solid (s) at 25°C.

We are given the mass of Na (s) used in the reaction (.560g) and the amount of heat evolved (13.8 kJ). We can use the molar mass of Na (22.99 g/mol) to calculate the moles of Na used in the reaction:

moles of Na = (.560g Na) / (22.99 g/mol Na) = 0.0243 mol Na

Since the reaction is at standard state conditions, we can use the definition of standard enthalpy of formation to set up an equation:

(delta Hf) = (heat evolved) / (moles of product formed)

Using the values we calculated, we can plug them in to solve for the standard enthalpy of formation of NaF:

(delta Hf) = (13.8 kJ) / (0.0243 mol) = -569 kJ/mol

Therefore, the standard enthalpy of formation of NaF is -569 kJ/mol.