When the oxide of generic metal M is heated at 25.0 °C, only a negligible amount of M is produced.
O2M(s) ----> M(s) +O2(g) delta G= 290.5 Kj/mol
When this reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous. What is the chemical equation of this coupled process? Show that the reaction is in equilibrium, include physical states, and represent graphite as C(s).
I got the coupled reaction is:
O2M(S)+C(S)>>M(S)+CO2(G)
ive been trying for the thermodynamic equillibrium constant which is the next part of the question:
I got ln(0.04193) and did e^-3.17 but its wrong. can anyone help?
I've just moved and don't know where my tables are but here is what you do.
MO2 ==> M + O2 dG = 290.5 kJ/mol
C + O2 ==> CO2 dG = look in the tables for dGf.
Add equation 1 to equation 2 to obtain the equation you want which is
MO2 + C ==> M + CO2 and dGrxn = dGf for eqn 1 + dGf for eqn 2.
Then dGrxn = -8.314(298)*lnK
thankyouu so much
To solve this problem, we can start by understanding the concept of coupling reactions and their equilibrium constant.
Coupling reactions involve combining two or more chemical reactions to create an overall reaction. In this case, the reaction of interest can be represented as:
O2M(s) + C(s) → M(s) + CO2(g)
To determine if this reaction is in equilibrium, we need to calculate the equilibrium constant using the Gibbs free energy change. We have the value of ΔG for the spontaneous reaction involving the conversion of O2M(s) to M(s) and O2(g), which is ΔG = 290.5 kJ/mol.
The equilibrium constant (K) can be calculated using the equation:
ΔG = -RT ln(K)
Where:
ΔG is the Gibbs free energy change
R is the gas constant (8.314 J/mol·K)
T is the temperature in Kelvin
K is the equilibrium constant
Since the given value of ΔG is in kJ/mol, we need to convert it to J/mol:
ΔG = 290.5 kJ/mol x (1000 J/1 kJ) = 290,500 J/mol
Next, we need to convert the temperature from Celsius to Kelvin:
T = 25.0 °C + 273.15 = 298.15 K
Plugging these values into the equation, we have:
290,500 J/mol = -(8.314 J/mol·K)(298.15 K) ln(K)
Now we can solve for ln(K):
ln(K) = (290,500 J/mol) / (8.314 J/mol·K)(298.15 K)
ln(K) ≈ 115.074
To find K, we can take the antilog of ln(K) using the exponential function, e:
K ≈ e^115.074
K ≈ 6.059 x 10^49
Therefore, the equilibrium constant for the coupled reaction O2M(s) + C(s) → M(s) + CO2(g) is approximately 6.059 x 10^49.