Nitrogen Dioxide , a pollutant in the atmosphere, can combine with water to form nitric acid. One of the possible reactions is shown here. Calculate delta ∆G° and Kp for this reaction at 25°C and comment on the spontaneity of the reaction. 3NO2(g)+H2O(l)--> 2 HNO3(aq)+ NO (g).
Temp Kp
170 K 3.8x 10^-3
180 K 0.34
190 K 18.4
200 K 681
To calculate the standard Gibbs free energy change (∆G°) and the equilibrium constant (Kp) for the given reaction at 25°C, we need the Kp values at different temperatures and use thermodynamic principles. Let's begin the calculation step by step:
1. Convert the given Kp values to Kc values:
The given Kp values are at different temperatures, but we need the Kc values (equilibrium constant in terms of concentrations) to calculate ∆G°. The relationship between Kp and Kc is given by the equation Kp = Kc(RT)^(∆n), where R is the gas constant and ∆n is the change in the number of moles of gas molecules. Since this equation relates Kp and Kc at a specific temperature, we will need to make use of the ideal gas law to convert Kp values to Kc values.
2. Convert Kp to Kc using the ideal gas law:
By rearranging the ideal gas law equation PV = nRT, we can write P = (n/V)RT, where P is pressure in atm, n/V is molar concentration, R is the gas constant, and T is the temperature in Kelvin. Applying this to the given reaction: 3NO2(g) + H2O(l) -> 2HNO3(aq) + NO(g), we observe that there are three moles of NO2 and one mole of H2O on the reactant side, while two moles of HNO3 and one mole of NO are produced. Therefore, ∆n = (2 + 1) - (3 + 1) = -1.
3. Calculating Kc values:
Using the equation Kp = Kc(RT)^(∆n), we can rearrange it to Kc = Kp / (RT)^(-∆n). Let's calculate the Kc values at each temperature provided:
- At 170 K: Kc = (3.8 x 10^(-3)) / (0.08206 x 170)^(-(-1)) = 1.2 x 10^(-3)
- At 180 K: Kc = (0.34) / (0.08206 x 180)^(-(-1)) = 1.9
- At 190 K: Kc = (18.4) / (0.08206 x 190)^(-(-1)) = 15
- At 200 K: Kc = (681) / (0.08206 x 200)^(-(-1)) = 126
4. Calculating ∆G°:
To calculate ∆G°, we can use the equation ∆G° = -RTln(Kc), where R is the gas constant and T is the temperature in Kelvin. Substituting the values:
- At 25°C (298 K): ∆G° = -(0.08206 x 298) x ln(15) = -6.45 kJ/mol (approximately)
5. Comment on spontaneity:
For a reaction to be spontaneous at a given temperature, ∆G° must be negative. In this case, ∆G° is approximately -6.45 kJ/mol at 25°C, indicating that the reaction is indeed spontaneous.
Therefore, the calculated values are:
∆G° = -6.45 kJ/mol
Kc = 15
And, based on the calculated ∆G°, we can conclude that the reaction is spontaneous.
dGorxn = (n*dGo products) - (dGo products)
dGo = -RT*lnK
Solve for K @ 298 and comment on spontaneity.