3CH4(g)->C3H8(g)+2H2(g)

Calculate change in G at 298 k if the reaction mixture consists of 41atm of CH4 , 0.010 atm of CH3, and 2.3×10−2 atm of H2.

To calculate the change in Gibbs free energy (ΔG) at 298 K, we can use the equation:

ΔG = ΔG° + RT ln(Q)

Where:
- ΔG is the change in Gibbs free energy
- ΔG° is the standard Gibbs free energy change for the reaction at standard conditions (1 atm and 298 K)
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (298 K)
- ln is the natural logarithm
- Q is the reaction quotient, which is the ratio of the partial pressures of the products to the partial pressures of the reactants.

Given the balanced equation: 3CH₄(g) → C₃H₈(g) + 2H₂(g)

First, we need to calculate the reaction quotient (Q):
Q = (P(C₃H₈) * P(H₂)²) / P(CH₄)³

Next, we need to calculate the values of the partial pressures using the given data:

P(CH₄) = 41 atm
P(C₃H₈) = 0.010 atm
P(H₂) = 2.3×10^(-2) atm

Now we can substitute the values into the equation to calculate ΔG:

ΔG = ΔG° + RT ln(Q)

Since the reaction takes place under standard conditions (1 atm), we assume that ΔG° = 0.

ΔG = 0 + (8.314 J/(mol·K)) * 298 K * ln([(0.010 atm)*(2.3×10^(-2) atm)²] / (41 atm)³)

Now we can calculate the values inside the natural logarithm:

ln([(0.010 atm)*(2.3×10^(-2) atm)²] / (41 atm)³) = ln(1.7668×10^(-10)) ≈ -23.307

Substituting this value back into the equation:

ΔG = (8.314 J/(mol·K)) * 298 K * (-23.307) ≈ -56,898 J/mol

Therefore, the change in Gibbs free energy (ΔG) for the given reaction mixture is approximately -56,898 J/mol at 298 K.

To calculate the change in Gibbs free energy (∆G) for the given reaction at 298 K, we need to use the equation:

∆G = ∑νGf(products) - ∑νGf(reactants)

Where:
- ∆G is the change in Gibbs free energy
- ν is the stoichiometric coefficient of the species in the reaction
- Gf is the standard molar Gibbs free energy of formation

First, let's calculate the ∆G for the products:

∆G(products) = ν[C3H8(g)] * Gf[C3H8(g)] + ν[H2(g)] * Gf[H2(g)]

Given:
ν[C3H8(g)] = 1
ν[H2(g)] = 2

We need to look up the standard molar Gibbs free energy of formation (Gf) for C3H8(g) and H2(g) at 298 K. Let's assume:
Gf[C3H8(g)] = 0 kJ/mol
Gf[H2(g)] = 0 kJ/mol (Assuming they are in their standard states)

Next, let's calculate ∆G for the reactant:

∆G(reactant) = ν[CH4(g)] * Gf[CH4(g)] + ν[CH3(g)] * Gf[CH3(g)] + ν[H2(g)] * Gf[H2(g)]

Given:
ν[CH4(g)] = -1
ν[CH3(g)] = 1
ν[H2(g)] = 2

Again, we need to look up the standard molar Gibbs free energy of formation (Gf) for CH4(g) and CH3(g) at 298 K. Let's assume:
Gf[CH4(g)] = 0 kJ/mol
Gf[CH3(g)] = 0 kJ/mol (Assuming they are in their standard states)

Now, let's substitute the values and calculate the change in Gibbs free energy (∆G):

∆G = (1 * 0) + (2 * 0) - (-1 * 0) + (1 * 0) + (2 * 0)

∆G = 0 kJ/mol

Basically, Hess's law requires that you manipulate the equations (either by multiplying or dividing by a factor) and add (or subtract) the equations, such that terms in the final combined equation cancel out on either side to leave you with the equation you are trying to end up with.

Whenver you operate on a single equation (say, multiply by 2), you need to multiply the value of delta H by 2.

For your problem above, you would need the value of delta h in the final problem. From what I can see, you would need to do the following:

1. Multiply the entire first equation by two (2) which would leave you with 2H2 (g) + O2 (g) -------> 2H2O and delta H = -571.6 kJ

2. Reverse the third equation, which would leave you with
3CO2 + 4H2O ----------> C3H8 + 5O2 and delta H = (the negative of whatever the omitted value is i.e. if it is -300, then it should be 300 and vice versa).

3. Add equation 2 to the results obtained from (1) and (2) above. You will see that species cancel out to give you
C3H4(g)+2H2(g) ----> C3H8(g)

the delta H value of this equation is then -571.6 + (-1937) + (********) kJ