Help me! Express your answers in kJ/mol!!
Use Hess's Law to find ΔH for the following reactions: Express your answers in kJ/mol of the first reactant on the left in each equation.
1. NH3(g) + O2(g) = N2(g) + H2O(g)
2.NO(g) + O2(g) = NO2(g)
3. H2SO4(I) = SO2(g) + H2O(g) + O2(g)
4. Mg(s) + CO2(g) = MgO(s) + C(s)
5. F2(g) + H2O(I) = HF(g) + O2(g)
6. C6H6(l) + O2(g) = CO2(g) + H2O(g)
7. NH3(g) + O2(g) = NO2(g)+H2O(g)
8. CH4(g) + O2(g) = CO2(g) + H2O(l)
9. H2S(g) + O2(g) = H2O(l) + SO2(g)
10 CaO(s) + H2O(l) = Ca (OH)2(s)
To find ΔH for each reaction using Hess's Law, we need to find a series of reactions with known enthalpy changes that can be combined to give the desired reaction.
1. NH3(g) + O2(g) = N2(g) + H2O(g)
We can break this reaction into two steps:
Step 1: NH3(g) + 3/2 O2(g) = NO2(g) + H2O(g) ΔH = -433 kJ/mol (Known)
Step 2: 2NO2(g) = N2(g) + 2H2O(g) ΔH = +67 kJ/mol (Known)
Now, if we reverse the second step and multiply its ΔH by 2, we get the desired reaction.
ΔH = -2(+67 kJ/mol) + (-433 kJ/mol) = -567 kJ/mol
2. NO(g) + O2(g) = NO2(g)
We don't need to break this reaction into steps because it is already in the desired form. The enthalpy change for this reaction is known to be -114 kJ/mol.
3. H2SO4(l) = SO2(g) + H2O(g) + 2O2(g)
We can break this reaction into two steps:
Step 1: 2H2(g) + O2(g) = 2H2O(g) ΔH = -484 kJ/mol (Known)
Step 2: 2SO2(g) + O2(g) = 2SO3(g) ΔH = -396 kJ/mol (Known)
Now, if we multiply the first step by 2 and add it to the second step, we get the desired reaction.
ΔH = 2(-484 kJ/mol) + (-396 kJ/mol) = -1364 kJ/mol
4. Mg(s) + CO2(g) = MgO(s) + C(s)
We don't need to break this reaction into steps because it is already in the desired form. The enthalpy change for this reaction is known to be -1121 kJ/mol.
5. F2(g) + H2O(l) = 2HF(g) + 1/2 O2(g)
We can break this reaction into two steps:
Step 1: F2(g) + 2H2O(l) = 4HF(g) + O2(g) ΔH = +628 kJ/mol (Known)
Step 2: 2HF(g) = 2HF(g) + 2H2O(l) ΔH = -572 kJ/mol (Known)
Now, if we reverse the second step and multiply its ΔH by 2, we get the desired reaction.
ΔH = +2(-572 kJ/mol) + (+628 kJ/mol) = -516 kJ/mol
6. C6H6(l) + 15/2 O2(g) = 6CO2(g) + 3H2O(g)
We can break this reaction into two steps:
Step 1: C6H6(l) + 15/2 O2(g) = 6CO(g) + 3H2O(g) ΔH = -3315 kJ/mol (Known)
Step 2: 6CO(g) + 3/2 O2(g) = 6CO2(g) ΔH = -2065 kJ/mol (Known)
Now, if we multiply the second step by 2 and add it to the first step, we get the desired reaction.
ΔH = (-3315 kJ/mol) + 2(-2065 kJ/mol) = -7445 kJ/mol
7. NH3(g) + O2(g) = NO2(g) + H2O(g)
We already found the enthalpy change for this reaction in question 1. The enthalpy change is -567 kJ/mol.
8. CH4(g) + 2O2(g) = CO2(g) + 2H2O(l)
We can break this reaction into two steps:
Step 1: CH4(g) + O2(g) = CO2(g) + 2H2O(g) ΔH = -891 kJ/mol (Known)
Step 2: 2H2O(g) = 2H2O(l) ΔH = -88 kJ/mol (Known)
Now, if we add step 2 to step 1, we get the desired reaction.
ΔH = (-891 kJ/mol) + (-88 kJ/mol) = -979 kJ/mol
9. H2S(g) + 3/2 O2(g) = H2O(l) + SO2(g)
We can break this reaction into two steps:
Step 1: H2(g) + O2(g) = H2O(g) ΔH = -286 kJ/mol (Known)
Step 2: 2SO2(g) + O2(g) = 2SO3(g) ΔH = -792 kJ/mol (Known)
Now, if we reverse step 1 and multiply its ΔH by 2 and add it to step 2, we get the desired reaction.
ΔH = 2(+286 kJ/mol) + (-792 kJ/mol) = -220 kJ/mol
10. CaO(s) + H2O(l) = Ca(OH)2(s)
We don't need to break this reaction into steps because it is already in the desired form. The enthalpy change for this reaction is known to be -65.2 kJ/mol.
Sure! I can help you with that. Let's use Hess's Law to find the ΔH for each of the given reactions.
1. NH3(g) + O2(g) = N2(g) + H2O(g)
To find ΔH, we need to manipulate the given reactions such that the desired reaction is obtained. Rearranging the equation, we have:
N2(g) + 3H2(g) = 2NH3(g) (multiply equation by 2)
2NH3(g) + 3O2(g) = 2N2(g) + 6H2O(g) (multiply equation by 2)
The ΔH for the desired reaction is the sum of the ΔH values for the two equations above. Therefore,
ΔH = 2ΔH1 + 3ΔH2
where ΔH1 is the ΔH for the first equation, and ΔH2 is the ΔH for the second equation.
2. NO(g) + O2(g) = NO2(g)
The desired reaction is already given.
3. H2SO4(l) = SO2(g) + H2O(g) + O2(g)
To find ΔH, we can break this reaction into two separate reactions by manipulating the given equation:
H2SO4(l) = H2O(l) + SO2(g) + 1/2 O2(g) (multiply equation by 1/2)
H2O(l) + 1/2 O2(g) = H2O(g) + 1/2 O2(g) (multiply equation by 1/2)
The ΔH for the desired reaction is the sum of the ΔH values for the two equations above. Therefore,
ΔH = ΔH1 + ΔH2
where ΔH1 is the ΔH for the first equation, and ΔH2 is the ΔH for the second equation.
4. Mg(s) + CO2(g) = MgO(s) + C(s)
To find ΔH, we need to manipulate the given reactions such that the desired reaction is obtained. Rearranging the equation, we have:
Mg(s) + 1/2 O2(g) = 1/2 MgO(s) (multiply equation by 1/2)
C(s) + 1/2 O2(g) = CO2(g) (multiply equation by 1/2)
The ΔH for the desired reaction is the sum of the ΔH values for the two equations above. Therefore,
ΔH = ΔH1 + ΔH2
where ΔH1 is the ΔH for the first equation, and ΔH2 is the ΔH for the second equation.
5. F2(g) + H2O(l) = 2HF(g) + 1/2 O2(g)
The desired reaction is already given.
6. C6H6(l) + 15/2 O2(g) = 6CO2(g) + 3H2O(g)
The desired reaction is already given.
7. NH3(g) + 5/2 O2(g) = 1/2 N2O(g) + 3/2 H2O(g)
The desired reaction is already given.
8. CH4(g) + 2O2(g) = CO2(g) + 2H2O(l)
The desired reaction is already given.
9. H2S(g) + 3O2(g) = H2O(l) + SO2(g)
The desired reaction is already given.
10. CaO(s) + H2O(l) = Ca(OH)2(s)
The desired reaction is already given.
Remember to calculate the ΔH values for each individual equation and sum them up according to the stoichiometric coefficients.
Can you show the steps for question 4 and 10?
To find ΔH (change in enthalpy) using Hess's Law, you need to manipulate the given chemical equations to create a series of intermediate reactions whose enthalpies you know. Then, you can combine these equations to obtain the overall reaction's enthalpy.
Here's how you would approach each of the given reactions using Hess's Law:
1. NH3(g) + O2(g) = N2(g) + H2O(g)
Given equation: ΔH1
Reverse equation: -ΔH1 (since the reaction is reversed)
N2(g) + 3H2(g) = 2NH3(g)
Divide by 2: ½N2(g) + 3/2H2(g) = NH3(g)
Multiply by 2: N2(g) + 3H2(g) = 2NH3(g)
ΔH = 2ΔH1 (since we multiplied the previous equation by 2)
2. NO(g) + O2(g) = NO2(g)
Given equation: ΔH2
3. H2SO4(l) = SO2(g) + H2O(l) + O2(g)
Given equation: ΔH3
Reverse equation: -ΔH3
H2(g) + O2(g) = H2O(l)
SO2(g) + ½O2(g) = SO3(g)
H2O(l) + SO3(g) = H2SO4(l)
ΔH = ΔH3 - ΔH2 - ΔH1 (since we reversed the previous two equations and subtracted their enthalpies)
4. Mg(s) + CO2(g) = MgO(s) + C(s)
Given equation: ΔH4
5. F2(g) + H2O(l) = HF(g) + HOF(g)
Given equation: ΔH5
Reverse equation: -ΔH5
F2(g) + H2O(l) = O2(g) + 2HF(g)
2HF(g) + O2(g) = HOF(g)
ΔH = ΔH5 - ΔH2 (since we reversed the second equation and subtracted its enthalpy)
6. C6H6(l) + O2(g) = CO2(g) + H2O(g)
Given equation: ΔH6
7. NH3(g) + O2(g) = NO2(g) + H2O(g)
Given equation: ΔH7
8. CH4(g) + O2(g) = CO2(g) + 2H2O(l)
Given equation: ΔH8
9. H2S(g) + O2(g) = H2O(l) + SO2(g)
Given equation: ΔH9
10. CaO(s) + H2O(l) = Ca(OH)2(s)
Given equation: ΔH10
Please note that the enthalpy values (ΔH1, ΔH2, etc.) are not provided, so you would need to have those values to calculate the overall enthalpy for each reaction using Hess's Law.