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) Well, if you pay close attention to the delta H value, you'll notice it's negative, which means the reaction is exothermic. It's getting hot in here, so take off all your cl...oh wait, that's not right.
(b) To calculate the amount of heat transferred, we need to use some stoichiometry. According to the balanced equation, 2 moles of Na produce -821.8 kJ of heat. So, let's find the moles of Na in 5.6 g and then use that to calculate the heat transferred.
The molar mass of Na is 22.99 g/mol. Therefore, the moles of Na in 5.6 g is:
Moles of Na = (5.6 g Na) / (22.99 g/mol Na)
Now, use this value to calculate the heat transferred:
Heat transferred = (Moles of Na) * (-821.8 kJ/ 2 mol Na)
(c) Ah, the production of NaCl! To determine the grams of NaCl produced during an enthalpy change of 16.5 kJ, we need to use the stoichiometry from the balanced equation once again. Based on the equation, 2 moles of Na produce -821.8 kJ of heat.
We can set up a proportion to solve for the moles of NaCl produced:
(Moles of NaCl) / (Moles of Na) = 2 moles NaCl / 2 moles Na
Now, use this proportion to calculate the moles of NaCl produced:
Moles of NaCl = (Moles of Na) * (2 moles NaCl / 2 moles Na)
Finally, convert moles to grams using the molar mass of NaCl to find the mass produced.
(d) Oh, look at that! We're decomposing NaCl now. To determine the heat absorbed, let's think about the balanced equation once again. The equation tells us that 2 moles of NaCl produce -821.8 kJ of heat.
To calculate the heat absorbed when 44.1 g of NaCl is decomposed, we need to use stoichiometry. First, let's find the moles of NaCl in 44.1 g:
Moles of NaCl = (44.1 g NaCl) / (molar mass NaCl)
Now, use this value to calculate the heat absorbed:
Heat absorbed = (Moles of NaCl) * (-821.8 kJ/ 2 mol NaCl)
And there you have it, the answers to all your heat-related questions!
(a) To determine if the reaction is exothermic or endothermic, we look at the sign of the enthalpy change (ΔH). In this case, the value of ΔH is -821.8 kJ. Since the ΔH is negative, the reaction is exothermic.
(b) To calculate the amount of heat transferred when 5.6 g of Na reacts at constant pressure, we need to use the stoichiometry of the reaction and the molar mass of Na.
First, calculate the moles of Na:
Molar mass of Na = 22.99 g/mol
Moles of Na = mass (g) / molar mass (g/mol) = 5.6 g / 22.99 g/mol = 0.2433 mol
Next, use the stoichiometry of the reaction to relate the moles of Na to the heat transferred:
2 moles of Na : -821.8 kJ
0.2433 mol of Na : x kJ
Using a proportion, we can solve for x:
0.2433 mol * (-821.8 kJ) / 2 mol = -99.82 kJ (rounded to two decimal places)
Therefore, the amount of heat transferred when 5.6 g of Na reacts at constant pressure is approximately -99.82 kJ.
(c) To determine the number of grams of NaCl produced during an enthalpy change of 16.5 kJ, we will use the stoichiometry of the reaction and the molar mass of NaCl.
The balanced equation gives us the stoichiometry:
2Na + Cl2 → 2NaCl
From this equation, we can see that 2 moles of NaCl are produced for every 2 moles of Na reacting. This means that the enthalpy change of -821.8 kJ is associated with the production of 2 moles of NaCl.
Now, let's calculate the number of moles of NaCl associated with an enthalpy change of 16.5 kJ using the proportion:
x mol of NaCl: 16.5 kJ
2 mol of NaCl: -821.8 kJ
x mol = 16.5 kJ * 2 mol of NaCl / -821.8 kJ
Simplifying gives us:
x mol ≈ 0.040 mol
Next, we can convert moles to grams using the molar mass of NaCl:
Molar mass of NaCl = 58.44 g/mol
Grams of NaCl = moles of NaCl * molar mass of NaCl
Grams of NaCl ≈ 0.040 mol * 58.44 g/mol ≈ 2.3376 g (rounded to four decimal places)
Therefore, approximately 2.3376 grams 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 at constant pressure, we will use the stoichiometry of the reaction and the molar mass of NaCl.
The balanced equation gives us the stoichiometry:
2NaCl → 2Na + Cl2
From this equation, we can see that 2 moles of NaCl are decomposed to produce 2 moles of Na and 1 mole of Cl2. This means that the enthalpy change of -821.8 kJ is associated with the decomposition of 2 moles of NaCl.
Now, let's calculate the number of moles of NaCl associated with an enthalpy change of -821.8 kJ using the proportion:
2 mol of NaCl : -821.8 kJ
x mol of NaCl : y kJ
x mol ≈ 44.1 g * (1 mol / 58.44 g/mol) ≈ 0.7547 mol
Using the proportion:
0.7547 mol * (-821.8 kJ) / 2 mol = -313.623 kJ (rounded to three decimal places)
Therefore, approximately -313.623 kJ of heat are absorbed when 44.1 g of NaCl is decomposed into Na and Cl2 at constant pressure. Note that the negative sign indicates that heat is absorbed.
To determine the answers to the questions, we need to use the given information, along with the concepts of enthalpy change and stoichiometry. Let's break down each question and calculate the results step by step.
(a) Is the reaction exothermic or endothermic?
To determine whether the reaction is exothermic or endothermic, we need to look at the sign of ΔH. In this case, ΔH is given as -821.8 kJ. Since the sign is negative, the reaction is exothermic.
(b) Calculate the amount of heat transferred when 5.6 g of Na reacts at constant pressure.
To solve this question, we need to use the molar mass of Na and the stoichiometry from the balanced equation.
1. Calculate the number of moles of Na:
Molar mass of Na = 22.99 g/mol
Number of moles of Na = mass / molar mass
Number of moles of Na = 5.6 g / 22.99 g/mol
2. Use stoichiometry to determine the heat transferred:
From the balanced equation, we know that 2 moles of Na produce -821.8 kJ of heat.
So, 1 mole of Na produces -821.8 kJ / 2 moles = -410.9 kJ.
To calculate the heat transferred for 5.6 g of Na, multiply the moles of Na by the heat transferred per mole:
Heat transferred = number of moles of Na * heat transferred per mole
Heat transferred = (5.6 g / 22.99 g/mol) * (-410.9 kJ)
(c) How many grams of NaCl are produced during an enthalpy change of 16.5 kJ?
To solve this question, we need to use stoichiometry and the heat transferred.
1. Use stoichiometry to relate heat transferred to the amount of NaCl produced:
From the balanced equation, we know that 2 moles of NaCl produce -821.8 kJ of heat.
So, 1 mole of NaCl produces -821.8 kJ / 2 moles = -410.9 kJ.
2. Calculate the number of moles of NaCl produced:
Moles of NaCl = heat transferred / heat transferred per mole
Moles of NaCl = 16.5 kJ / (-410.9 kJ/mol)
3. Calculate the mass of NaCl produced:
Mass of NaCl = moles of NaCl * molar mass of NaCl
(d) How many kilojoules of heat are absorbed when 44.1 g of NaCl is decomposed into Na and Cl2 at constant pressure?
To solve this question, we can use the same approach as in part (b) but in reverse.
1. Calculate the number of moles of NaCl:
Molar mass of NaCl = 58.44 g/mol
Number of moles of NaCl = mass / molar mass
Number of moles of NaCl = 44.1 g / 58.44 g/mol
2. Use stoichiometry to determine the heat absorbed:
From the balanced equation, we know that 2 moles of NaCl produce -821.8 kJ of heat.
So, 1 mole of NaCl absorbs -821.8 kJ / 2 moles = -410.9 kJ.
To calculate the heat absorbed for 44.1 g of NaCl, multiply the moles of NaCl by the heat absorbed per mole:
Heat absorbed = number of moles of NaCl * heat absorbed per mole
Heat absorbed = (44.1 g / 58.44 g/mol) * (-410.9 kJ)
By following these calculations, you can determine the answers to all four questions.