Question

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.


4. Mg(s) + CO2(g) = MgO(s) + C(s)
10 CaO(s) + H2O(l) = Ca (OH)2(s)

Answer 1

are you sure this answers are correct?

Answer 2

To find ΔH for the given reactions using Hess's Law, we need to manipulate and combine other reactions that have known enthalpy changes.

4. Mg(s) + CO2(g) = MgO(s) + C(s)

First, we need to find the reactions that have known enthalpy changes that can combine to give the desired reaction. We know the following reactions:

1. C(s) + O2(g) = CO2(g) ΔH = -393.5 kJ/mol
2. Mg(s) + 1/2 O2(g) = MgO(s) ΔH = -601.8 kJ/mol

Now, we manipulate the equations to get the desired reaction:

3. C(s) + O2(g) = CO2(g)
Multiply equation 3 by 1/2 to match the coefficient of CO2 in the desired reaction:

4. 1/2 C(s) + 1/2 O2(g) = 1/2 CO2(g)

Now, we can combine equations 2 and 4 to get the desired reaction:

5. Mg(s) + 1/2 C(s) + 1/2 O2(g) = MgO(s) + 1/2 CO2(g)

The enthalpy change for equation 5 can be calculated by summing the enthalpy changes of equations 2 and 4:

ΔH = ΔH2 + ΔH4
ΔH = -601.8 kJ/mol + [-393.5 kJ/mol] (Note the negative sign for equation 4)

Now, plug in the values:

ΔH = -601.8 kJ/mol + (-393.5 kJ/mol)
ΔH = -995.3 kJ/mol

Therefore, ΔH for the reaction Mg(s) + CO2(g) = MgO(s) + C(s) is -995.3 kJ/mol of Mg(s) (the first reactant on the left).

10. CaO(s) + H2O(l) = Ca(OH)2(s)

Similar to the previous reaction, we need to manipulate other reactions to obtain the desired reaction. We know the following reaction:

1. H2(g) + 1/2 O2(g) = H2O(l) ΔH = -285.8 kJ/mol

Now, we manipulate the equation to get the desired reaction:

2. Ca(OH)2(s) = CaO(s) + H2O(l)

Now, we can combine equations 1 and 2 to get the desired reaction:

3. H2(g) + 1/2 O2(g) + Ca(OH)2(s) = CaO(s) + H2O(l)

The enthalpy change for equation 3 can be calculated by summing the enthalpy changes of equations 1 and 2:

ΔH = ΔH1 + ΔH2
ΔH = -285.8 kJ/mol + [unknown ΔH for Ca(OH)2(s) = CaO(s) + H2O(l)]

Now, we need to find the value of ΔH for the reaction 2 (the unknown).

Therefore, ΔH for the reaction CaO(s) + H2O(l) = Ca(OH)2(s) is unknown.

Answer 3

Apologies for the confusion in my previous responses. The calculations for both reactions were incorrect. Let's correct them:

4. Mg(s) + CO2(g) = MgO(s) + C(s)

Given reactions:
1. C(s) + O2(g) = CO2(g) ΔH = -393.5 kJ/mol
2. Mg(s) + 1/2 O2(g) = MgO(s) ΔH = -601.8 kJ/mol

To obtain the desired reaction, we manipulate the given reactions as follows:

Multiply reaction 1 by 1/2 to match the coefficient of CO2 in the desired reaction:

3. 1/2 C(s) + 1/2 O2(g) = 1/2 CO2(g)

Now, we combine equations 2 and 3 to get the desired reaction:

4. Mg(s) + 1/2 C(s) + 1/2 O2(g) = MgO(s) + 1/2 CO2(g)

The enthalpy change for equation 4 can be calculated by summing the enthalpy changes of equations 2 and 3:

ΔH = ΔH2 + ΔH3

ΔH = -601.8 kJ/mol + [-1/2 * (-393.5 kJ/mol)] (Note the negative sign for equation 3)

Now, plug in the values:

ΔH = -601.8 kJ/mol + [1/2 * (-393.5 kJ/mol)]
ΔH = -601.8 kJ/mol + (-196.8 kJ/mol)
ΔH = -798.6 kJ/mol

Therefore, ΔH for the reaction Mg(s) + CO2(g) = MgO(s) + C(s) is -798.6 kJ/mol of Mg(s) (the first reactant on the left).

10. CaO(s) + H2O(l) = Ca(OH)2(s)

Given reaction:
1. H2(g) + 1/2 O2(g) = H2O(l) ΔH = -285.8 kJ/mol

In this case, the given reaction already matches the desired reaction. So the ΔH value for the given reaction (ΔH = -285.8 kJ/mol) is equal to the ΔH for the reaction CaO(s) + H2O(l) = Ca(OH)2(s).

Therefore, ΔH for the reaction CaO(s) + H2O(l) = Ca(OH)2(s) is -285.8 kJ/mol of CaO(s) (the first reactant on the left).