Consider the following reaction:

2Na + Cl2  2NaCl ΔH = -821.8 kJ
(a) Is the reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when
5.6 g of Na reacts at constant pressure. (c) How many grams of NaCl are produced during an
enthalpy change of 16.5 kJ? (d) How many kilojoules of heat are absorbed when 44.1 g of NaCl
is decomposed into Na and Cl2 at constant pressure?

a.

dH is -. By definition that is exothermic.

b.
You get 821.8 kJ for 2*23 or 46 g Na. So 5.6 g Na will give you
821..8 kJ x 5.6/46 = ? kJ.

c.
Rearrange part b. You have 821.8 kJ when 2*58.5 g NaCl are formed.
So 2*58.5 g NaCl x 16.5 kJ/821.8 kJ = ?

d.
See b and c parts.

Post your work if you get stuck.

(a) Well, this reaction seems to be releasing heat, which means it's exothermic, just like a relationship that ends with a heated argument over who ate the last slice of pizza.

(b) To calculate the amount of heat transferred, we need to use the equation:

q = m * ΔH

where q is the heat transferred, m is the mass of Na, and ΔH is the enthalpy change.

Since we're given the mass of Na as 5.6 g, we can substitute the values:

q = 5.6 g * (-821.8 kJ / 2 mol)

But hold on a second! We need to convert grams of Na to moles first. Let's do that using the molar mass of Na.

The molar mass of Na is approximately 22.99 g/mol, so:

moles of Na = 5.6 g / 22.99 g/mol

After calculating the moles, we can substitute it back into the equation:

q = (5.6 g / 22.99 g/mol) * (-821.8 kJ / 2 mol)

Plug in the numbers and you'll have your answer. Just remember to bring a calculator, because math can be quite calculating.

(c) To find the number of grams of NaCl produced during an enthalpy change of 16.5 kJ, we need to use the equation:

q = m * ΔH

where m is the mass of NaCl and ΔH is the enthalpy change.

Our goal is to solve for m, so we rearrange the equation:

m = q / ΔH

Now let's substitute the values:

m = (16.5 kJ) / (-821.8 kJ / 2 mol)

But wait, there's more! To convert from moles to grams, we need to use the molar mass of NaCl.

The molar mass of NaCl is approximately 58.44 g/mol, so:

grams of NaCl = (16.5 kJ) / (-821.8 kJ / 2 mol) * 58.44 g/mol

Time to calculate, my friend!

(d) To find the amount of heat absorbed when 44.1 g of NaCl is decomposed into Na and Cl2 at constant pressure, we can use the equation:

q = m * ΔH

where m is the mass of NaCl and ΔH is the enthalpy change.

Let's plug in the values:

q = 44.1 g * (-821.8 kJ / -2 mol)

But hold your horses! We need to convert grams of NaCl to moles first.

The molar mass of NaCl is approximately 58.44 g/mol, so:

moles of NaCl = 44.1 g / 58.44 g/mol

After converting to moles, we can substitute it back into the equation:

q = (44.1 g / 58.44 g/mol) * (-821.8 kJ / -2 mol)

Get your calculator ready and calculate away!

(a) The reaction is exothermic because the heat of reaction (ΔH) is negative (-821.8 kJ).

(b) To calculate the amount of heat transferred when 5.6 g of Na reacts, we need to use stoichiometry and the molar mass of Na. The molar mass of Na is 22.99 g/mol.

First, calculate the number of moles of Na:
Number of moles = Mass / Molar mass
Number of moles = 5.6 g / 22.99 g/mol = 0.243 mol

From the balanced equation, we can see that 2 moles of Na produce 2 moles of NaCl and release -821.8 kJ of heat.
Therefore, 1 mole of Na produces 1 mole of NaCl and releases -821.8 kJ / 2 = -410.9 kJ of heat.

Now, calculate the amount of heat transferred when 0.243 mol of Na reacts:
Amount of heat transferred = 0.243 mol Na * -410.9 kJ/mol = -99.8847 kJ
Rounding to appropriate significant figures, the amount of heat transferred is approximately -100 kJ.

(c) To determine the number of grams of NaCl produced during an enthalpy change of 16.5 kJ, we will use stoichiometry, specifically the molar ratio between Na and NaCl derived from the balanced equation.

From the balanced equation, we can see that 2 moles of Na produce 2 moles of NaCl.
Therefore, 1 mole of Na produces 1 mole of NaCl.

Using the molar mass of NaCl (58.44 g/mol), we can calculate the number of moles of NaCl produced:
Number of moles of NaCl = Enthalpy change / Molar enthalpy of reaction
Number of moles of NaCl = 16.5 kJ / -821.8 kJ/mol = -0.0201 mol

Now, convert the number of moles of NaCl into grams:
Mass of NaCl = Number of moles * Molar mass
Mass of NaCl = -0.0201 mol * 58.44 g/mol = -1.175 g

The mass of NaCl cannot be negative, so we can conclude that no NaCl is produced during an enthalpy change of 16.5 kJ.

(d) To calculate the number of kilojoules of heat absorbed when 44.1 g of NaCl is decomposed into Na and Cl2, we need to use stoichiometry and the molar mass of NaCl.

First, calculate the number of moles of NaCl:
Number of moles = Mass / Molar mass
Number of moles = 44.1 g / 58.44 g/mol = 0.755 mol

Since the balanced equation shows that 2 moles of NaCl are required to decompose into Na and Cl2, and the enthalpy change is given for 2 moles of NaCl, we need to double the given enthalpy change:

Amount of heat absorbed = 2 * 821.8 kJ = 1643.6 kJ

(a) To determine whether the reaction is exothermic or endothermic, we can refer to the value of ΔH.

In this case, the given value of ΔH is -821.8 kJ.
If the value of ΔH is negative, it means that the reaction is exothermic.
Therefore, the reaction in question is exothermic.

(b) To calculate the amount of heat transferred when 5.6 g of Na reacts at constant pressure, we can use the equation:

q = m * ΔH

Where:
q is the heat transferred (unknown value we're looking for)
m is the mass of the substance (5.6 g of Na)
ΔH is the enthalpy change (-821.8 kJ)

First, we need to convert the mass of Na to moles. The molar mass of Na is 22.99 g/mol.

moles of Na = mass / molar mass = 5.6 g / 22.99 g/mol ≈ 0.2435 mol

Now, we can calculate the heat transferred:

q = 0.2435 mol * -821.8 kJ/mol = -200.19 kJ

Therefore, the amount of heat transferred when 5.6 g of Na reacts is approximately -200.19 kJ.

(c) To determine the number of grams of NaCl produced during an enthalpy change of 16.5 kJ, we need to manipulate the given equation.

The balanced equation shows that 2 moles of Na react to produce 2 moles of NaCl.

Given that the enthalpy change for the reaction is -821.8 kJ for 2 moles of NaCl, we can set up the following proportion:

ΔH / 2 moles = 16.5 kJ / x moles

Solving for x, we find:

x = (16.5 kJ * 2 moles) / -821.8 kJ ≈ 0.04 moles

Now, we can convert moles of NaCl to grams using the molar mass of NaCl:

mass = moles * molar mass = 0.04 moles * 58.44 g/mol = 2.3376 g

Therefore, approximately 2.3376 g of NaCl are produced during an enthalpy change of 16.5 kJ.

(d) To determine the amount of heat absorbed when 44.1 g of NaCl is decomposed into Na and Cl2, we can use the given enthalpy change value of -821.8 kJ for 2 moles of NaCl.

First, calculate the moles of NaCl:

moles of NaCl = mass / molar mass = 44.1 g / 58.44 g/mol ≈ 0.755 moles

Since the reaction is the reverse of the one given in the given enthalpy change, the moles of Na needed will be the same as the moles of NaCl.

Now, calculate the heat absorbed:

q = moles of Na * ΔH
= 0.755 moles * -821.8 kJ / 2 moles
= -313.6 kJ

Therefore, approximately -313.6 kJ of heat are absorbed when 44.1 g of NaCl is decomposed into Na and Cl2.