a) Is the reaction shown below spontaneous under standard conditions at 25⁰C?
CH3OH (g) ----> CO (g) + 2H2 (g)
b) Will this reaction be spontaneous at 25⁰C given the following amounts of materials?
CH3OH at 1.20 atm CO at 0.080 atm H2 at 0.020 atm
To determine if a reaction is spontaneous under standard conditions at 25⁰C, we can calculate the standard Gibbs free energy change (∆G°) using the equation:
∆G° = ∆H° - T∆S°
where ∆H° is the standard enthalpy change and ∆S° is the standard entropy change. If ∆G° is negative, the reaction is spontaneous.
However, to determine spontaneity based on given amounts of materials, we need to calculate the reaction quotient (Q) and compare it to the equilibrium constant (K). The reaction quotient is calculated using the equation:
Q = (P(CO) * P(H2)^2) / P(CH3OH)
where P(CH3OH), P(CO), and P(H2) are the partial pressures of CH3OH, CO, and H2, respectively.
If Q is less than K, the reaction proceeds forward and is spontaneous. If Q is greater than K, the reaction proceeds in the reverse direction and is not spontaneous. If Q is equal to K, the reaction is at equilibrium and does not spontaneously proceed in either direction.
Now, let's calculate Q for the given reaction with the given partial pressures:
Q = (0.080 * (0.020)^2) / 1.20
Q = 0.00008 / 1.20
Q = 6.67 x 10^-5 atm
To determine the spontaneity of the reaction, we need to compare Q to the equilibrium constant K. Unfortunately, the equilibrium constant K was not provided. Therefore, we cannot determine the spontaneity of the reaction based on the given amounts of materials.
To determine if a reaction is spontaneous under standard conditions at 25⁰C, we need to calculate the change in Gibbs free energy (∆G) using the equation:
∆G = ∆G° + RTln(Q)
Where:
∆G = Change in Gibbs free energy
∆G° = Standard Gibbs free energy change
R = Gas constant (8.314 J/mol·K)
T = Temperature in Kelvin (25⁰C + 273 = 298 K)
ln(Q) = Natural logarithm of the reaction quotient
a) To determine if the reaction is spontaneous under standard conditions, we need the standard Gibbs free energy change (∆G°) for the reaction. If ∆G° is negative, the reaction is spontaneous.
b) To determine if the reaction is spontaneous given the specified amounts of materials, we will use the reaction quotient (Q) to calculate ∆G. If ∆G is negative, the reaction will be spontaneous.
Now let's calculate each part:
a) Spontaneity under standard conditions:
To determine the standard Gibbs free energy change, we need to know the standard Gibbs free energy of formation (ΔG°f) for each compound.
Using the standard Gibbs free energy of formation values:
ΔG°f(CH3OH) = -166.3 kJ/mol
ΔG°f(CO) = -137.1 kJ/mol
ΔG°f(H2) = 0 kJ/mol
ΔG° = ΣnΔG°f(products) - ΣnΔG°f(reactants)
ΔG° = [ΔG°f(CO) + 2ΔG°f(H2)] - [ΔG°f(CH3OH)]
ΔG° = [-137.1 + 2(0)] - [-166.3]
ΔG° = -137.1 + 166.3
ΔG° = 29.2 kJ/mol
Since ΔG° is positive (29.2 kJ/mol), the reaction is not spontaneous under standard conditions at 25⁰C.
b) Spontaneity given the specified amounts of materials:
To determine if the reaction is spontaneous, we need to calculate the reaction quotient (Q) and substitute it into the equation:
Q = [CO]^a * [H2]^b / [CH3OH]^c
Given the partial pressures:
[CO] = 0.080 atm
[H2] = 0.020 atm
[CH3OH] = 1.20 atm
Substitute the values into the equation and calculate Q:
Q = (0.080)^1 * (0.020)^2 / (1.20)^1
Q = 0.00032 / 1.20
Q = 0.000267
Now substitute Q into the equation to calculate ΔG:
∆G = ∆G° + RTln(Q)
∆G = 29.2 kJ/mol + (8.314 J/mol·K)(298 K)ln(0.000267)
∆G = 29.2 kJ/mol + (8.314 J/mol)(298 K)(-8.2)
∆G = 29.2 kJ/mol - 20.5 kJ/mol
∆G = 8.7 kJ/mol
Since ΔG is positive (8.7 kJ/mol), the reaction is not spontaneous at 25⁰C given the specified amounts of materials.