Consider the reaction 3CH_4(g)rightarrow C_3H_8(g)+ 2H_2(g).

Calculate delta G at 298 K if the reaction mixture consists of 40atm of CH_4, 0.015atm of C_3H_8, and 2.3×10^−2atm of H_2.

Do you have an answer? Have you tried

dG = -RT*lnQ

i got 6605kj but i think its too much.

To calculate the standard Gibbs free energy change, ΔG°, for a reaction, we can use the equation:

ΔG° = ΔG°f (products) - ΔG°f (reactants)

where ΔG°f represents the standard Gibbs free energy of formation.

First, we need to determine the standard Gibbs free energy of formation for each compound involved in the reaction. The standard Gibbs free energy of formation for a compound is the change in Gibbs free energy when the compound is formed from its elements in their standard states.

Next, we can use the standard Gibbs free energy of formation values to calculate ΔG°:

ΔG° = (nC3H8 x ΔG°f (C3H8)) + (nH2 x ΔG°f (H2)) - (nCH4 x ΔG°f (CH4))

where n represents the stoichiometric coefficients of the compounds in the balanced chemical equation.

Now, let's find the standard Gibbs free energy of formation values for each compound:

ΔG°f (C3H8) = -103.85 kJ/mol
ΔG°f (H2) = 0 kJ/mol
ΔG°f (CH4) = -50.81 kJ/mol

Using the given partial pressures of the reactants, we can convert them to concentrations using the ideal gas law:

Concentration (in mol/L) = Pressure (in atm) / (R x Temperature (in Kelvin))

where R is the ideal gas constant (0.0821 L x atm / K x mol).

Converting the given partial pressures to concentrations:

[C3H8] = 0.015 atm / (0.0821 L x atm / K x mol x 298 K) = 0.000724 mol/L
[H2] = 2.3x10^-2 atm / (0.0821 L x atm / K x mol x 298 K) = 0.001113 mol/L
[CH4] = 40 atm / (0.0821 L x atm / K x mol x 298 K) = 1.9385 mol/L

Now, plugging the values into the ΔG° equation:

ΔG° = (1 x ΔG°f (C3H8)) + (2 x ΔG°f (H2)) - (3 x ΔG°f (CH4))
= (1 x -103.85 kJ/mol) + (2 x 0 kJ/mol) - (3 x -50.81 kJ/mol)
= -103.85 kJ/mol + 0 kJ/mol + 152.43 kJ/mol
= 48.58 kJ/mol

Therefore, the standard Gibbs free energy change for the reaction at 298 K is 48.58 kJ/mol.