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.
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)
2 answers
Can you show the steps for question 4 and 10?
3 answers