Scientific Understanding

**HOW DO WE CALCULATE ****Δ**** G (First Way)?**

There are two different ways to calculate **Δ** G (Delta G). The first is look up the **Δ** G values on a Gibbs Free Energy Table (DELTA G) and then take the **Δ** G of the products minus the **Δ** G of the reactants. This is the same technique you used in the how to calculate **Δ** H section only you are going to use values found in a slightly different table (Gibbs Free Energy / Spontaneity Table). The equation for the **Δ** G is below:

Energy of products – Energy of reactants = **Δ** G

**VIDEO Calculate ****Δ**** G (DELTA G) Demonstrated Example 1**: Use the balanced chemical equation below and calculate its **Δ** G. (Use this link look up the **Δ** G_{f} values)

CH_{4(g)} + 2 O_{2(g)} —-> CO_{2(g)} + 2 H_{2}O_{(g)}

What is the energy of the molecules of the products?

Answer: CO_{2(g)} = -390 kJ/mol H_{2}O_{(g)} = -230 kJ/mol

Do any of the molecules in the products have any coefficients?

Answer: Yes, H_{2}O_{(g)} 2 H_{2}O_{(g)} = 2 * (-230 kJ/mol)

What is the energy of the molecules of the reactants?

Answer: CH_{4(g)} = -50 kJ/mol O_{2(g)} = 0

Do any of the molecules in the reactants have any coefficients?

Answer: Yes, O_{2(g)} 2 O_{2(g)} = 2 * (0)

What is the general formula for **Δ** G?

Answer: Energy of products – Energy of reactants = **Δ** G

Modify the **Δ** H equation for this chemical equation.

(CO_{2(g)} + 2 H_{2}O_{(g)}) – (CH_{4(g)} + 2 O_{2(g)}) = **Δ** G

Fill in the numbers for this specific equation.

(-390 + 2 * (-230)) – (-50 + 2 * (0)) = **Δ** G

Solve for **Δ** H

COMPLETE ANSWER: **Δ** G = -800 kJ/mol

**VIDEO Calculate ****Δ**** G (DELTA G) Demonstrated Example 2**: Use the **Δ** G and balanced chemical equation below and calculate the **Δ** G_{f} of H_{2}Ba_{(s)}. (Use this link look up the **Δ** G_{f} values)

2 NaH_{(g)} + BaCl_{2(s)} —-> H_{2}Ba_{(s)} + 2 NaCl_{(s)}**Δ** G = -480 kJ/mol

What is the energy of the molecules of the products?

Answer: H_{2}Ba_{(s)} = X kJ/mol NaCl_{(s)} = -384 kJ/mol

Do any of the molecules in the products have any coefficients?

Answer: Yes, NaCl_{(s)} 2 NaCl_{(s)} = 2 * (-384 kJ/mol)

What is the energy of the molecules of the reactants?

Answer: NaH_{(g)} = -34 kJ/mol BaCl_{2(s)} = -790 kJ/mol

Do any of the molecules in the reactants have any coefficients?

Answer: Yes, NaH_{(g)} 2 NaH_{(g)} = 2 * (-34 kJ/mol)

What is the general formula for **Δ** G?

Answer: Energy of products – Energy of reactants = **Δ** G

Modify the **Δ** G equation for this chemical equation.

(H_{2}Ba_{(s)} + 2 NaCl_{(s)}) – (2 NaH_{(g)} + BaCl_{2(s)}) = **Δ** G

Fill in the numbers for this specific equation.

( X + 2 * (-380)) – (2 * (-34) + -790) = – 480

Solve for X

COMPLETE ANSWER: **Δ** G_{f} of H_{2}Ba_{(s)} = -578 kJ/mol

**PRACTICE PROBLEMS**: Calculate the **Δ** G or the **Δ** G_{f} as needed. (Use this link look up the **Δ** G_{f})

Use the balanced chemical equation below and calculate its **Δ** G.

2 H_{2}O_{(g)} —-> 2 H_{2(g)} + O_{2(g)}

Answer: **Δ** G = -460 kJ/mol

Use the balanced chemical equation below and calculate its **Δ** G.

H^{+1}_{(aq)} + F ^{-1}_{(aq)} —-> HF_{(g)}

Answer: **Δ** G = +10 kJ/mol