1. The value of ΔH° for the reaction below is -6535 kJ.
_________ kJ of heat are released in the combustion of 16.0 g C6H6 (l).
2C6H6 (l) + 15O2 (g) --> 12CO2 (g) + 6H20 (l)
3. Given the following reactions
CaCO3 (s) --> CaO (s) + CO2 (g) ΔH = 178.1 kJ
C (s,graphite) + O2 (g) --> CO2 (g) ΔH = -393.5 kJ
the enthalpy of the reaction
CaCO3 (s) --> CaO (s) + C (s,graphite) + O2 (g)
is ________ kJ.
1. First, we need to find the moles of C6H6 in 16.0 g.
Moles = (16.0 g) / (78.11 g/mol) = 0.2048 moles
Since the given reaction involves 2 moles of C6H6, we need to calculate the amount of heat released per mole of C6H6:
ΔH per mole of C6H6 = (-6535 kJ) / 2 = -3267.5 kJ/mol
Now, we can calculate the heat released during the combustion of 0.2048 moles of C6H6:
Heat released = (0.2048 moles) x (-3267.5 kJ/mol) = -669.0 kJ
So, 669.0 kJ of heat are released in the combustion of 16.0 g C6H6 (l).
3. To find the enthalpy of the desired reaction, we can subtract the ΔH values of the given reactions:
ΔH (desired reaction) = 178.1 kJ - (-393.5 kJ) = 178.1 kJ + 393.5 kJ = 571.6 kJ
So, the enthalpy of the reaction CaCO3 (s) --> CaO (s) + C (s,graphite) + O2 (g) is 571.6 kJ.
To solve these problems, we can use the concept of Hess's Law, which states that the enthalpy change of a reaction is independent of the pathway between the initial and final states. This means that we can manipulate and combine known reactions to determine the enthalpy change of a target reaction.
1. The given reaction is:
2C6H6 (l) + 15O2 (g) --> 12CO2 (g) + 6H20 (l)
To find the heat released in the combustion of 16.0 g of C6H6, we need to use the molar mass of C6H6 and convert it to moles.
The molar mass of C6H6 is:
(6 * atomic mass of carbon) + (6 * atomic mass of hydrogen)
= (6 * 12.01 g/mol) + (6 * 1.01 g/mol)
= 78.1 g/mol
Now let's calculate the moles of C6H6:
moles = mass / molar mass
moles = 16.0 g / 78.1 g/mol
moles = 0.205 mol
Using stoichiometry, we can relate the moles of C6H6 to the heat released in the reaction.
From the given reaction, we can see that for every 2 moles of C6H6 combusted, 6535 kJ of heat is released (ΔH° = -6535 kJ).
So, for 0.205 mol of C6H6 combusted, the heat released can be calculated as:
heat released = (0.205 mol C6H6 / 2 mol C6H6) * 6535 kJ
heat released = 668.4175 kJ
Therefore, 668.4175 kJ of heat are released in the combustion of 16.0 g C6H6 (l).
3. The target reaction is:
CaCO3 (s) --> CaO (s) + C (s,graphite) + O2 (g)
We need to use the given reactions to manipulate and combine them to obtain the target reaction.
First, let's reverse the first given reaction and change the sign of ΔH:
CaO (s) + CO2 (g) --> CaCO3 (s) ΔH = -178.1 kJ
Second, let's multiply the second given reaction by 1/2 to balance the number of moles of CO2:
1/2 C (s,graphite) + O2 (g) --> 1/2 CO2 (g) ΔH = 1/2 (-393.5 kJ) = -196.8 kJ
Now, let's add these two reactions together to obtain the target reaction:
CaO (s) + CO2 (g) + 1/2 C (s,graphite) + O2 (g) --> CaCO3 (s) + 1/2 CO2 (g) + 1/2 C (s,graphite) + O2 (g)
The CO2 and C (s,graphite) cancel out on both sides of the equation, leaving only:
CaO (s) + O2 (g) --> CaCO3 (s)
The ΔH for the target reaction is the sum of the ΔH values of the added reactions:
ΔH = (-178.1 kJ) + (-196.8 kJ)
ΔH = -374.9 kJ
Therefore, the enthalpy of the reaction CaCO3 (s) --> CaO (s) + C (s,graphite) + O2 (g) is -374.9 kJ.
To find the answers to these questions, we need to use the concept of Hess's Law and the principle of conservation of energy.
1. In this question, we are given the value of ΔH° for the reaction and asked to determine the amount of heat released in the combustion of 16.0 g of C6H6 (liquid benzene). To find the amount of heat released, we need to use the equation:
Q = ΔH° * n
Where Q is the heat released, ΔH° is the enthalpy change per mole of reaction, and n is the number of moles of the substance involved in the reaction.
To calculate the number of moles of C6H6, we use its molar mass:
Molar mass of C6H6 = (6 * Atomic mass of C) + (6 * Atomic mass of H)
= (6 * 12.01 g/mol) + (6 * 1.01 g/mol)
= 78.11 g/mol
Number of moles of C6H6 = Mass / Molar mass
= 16.0 g / 78.11 g/mol
= 0.205 mol
Now, we can substitute these values into the equation:
Q = (-6535 kJ) * (0.205 mol)
= -1338.9 kJ (rounded to one decimal place)
Therefore, approximately 1338.9 kJ of heat are released in the combustion of 16.0 g of C6H6.
2. This question requires the use of Hess's Law and the knowledge of enthalpy changes for different reactions.
Given the two reactions:
1) CaCO3 (s) --> CaO (s) + CO2 (g) ΔH = 178.1 kJ
2) C (s, graphite) + O2 (g) --> CO2 (g) ΔH = -393.5 kJ
To determine the enthalpy of the reaction:
CaCO3 (s) --> CaO (s) + C (s, graphite) + O2 (g)
We need to apply the following steps:
Step 1: Reverse the second reaction by switching the sign of the enthalpy change:
C (s, graphite) + O2 (g) <-- CO2 (g) ΔH = 393.5 kJ
Step 2: Multiply the first reaction by 1 to match the number of moles of CaO and CO2 in the desired reaction:
CaCO3 (s) --> CaO (s) + CO2 (g) ΔH = 178.1 kJ
Step 3: Add the equations together to cancel out the CO2 term:
CaCO3 (s) + C (s, graphite) + O2 (g) --> CaO (s) + C (s, graphite) + O2 (g) ΔH = (178.1 kJ) + (-393.5 kJ)
Now, we can simplify the equation:
CaCO3 (s) --> CaO (s) + C (s, graphite) + O2 (g) ΔH = -215.4 kJ
Therefore, the enthalpy of the reaction is -215.4 kJ.