dG for the formation of Hi(g) from its gaseous elements is -20.2 kj/mol at 500k. when the partial pressure of HI is 10 atm, and I2 0.001 atm, what must the partial pressure of hydrogen be at this temperature to reduce the magnitude of dG for the reaction to 0. H2(g)+I2(g)---->2HI(g) answer: 775 atm.
To solve this problem, we can use the equation for Gibbs free energy, which is given by:
ΔG = ΔG° + RT ln(Q)
Where:
ΔG is the change in Gibbs free energy,
ΔG° is the standard Gibbs free energy change,
R is the gas constant (8.314 J/(mol·K) or 0.0821 L·atm/(mol·K)),
T is the temperature in Kelvin,
Q is the reaction quotient.
Given:
ΔG = 0 kJ/mol (we want to reduce the magnitude of ΔG to zero),
ΔG° = -20.2 kJ/mol (standard Gibbs free energy change),
T = 500 K,
P(HI) = 10 atm,
P(I2) = 0.001 atm,
P(H2) = ? (unknown partial pressure of hydrogen).
First, let's convert the given values to the appropriate units:
ΔG° = -20.2 kJ/mol = -20,200 J/mol
R = 0.0821 L·atm/(mol·K)
T = 500 K
P(HI) = 10 atm
P(I2) = 0.001 atm
Next, let's calculate the reaction quotient (Q):
Q = [HI]^2 / ([H2] * [I2])
Since we want to reduce the magnitude of ΔG to zero, we can set ΔG equal to zero:
0 = ΔG° + RT ln(Q)
Substituting the values into the equation:
0 = -20,200 J/mol + (0.0821 L·atm/(mol·K)) * (500 K) * ln(Q)
Now we can solve for Q:
ln(Q) = (20,200 J/mol) / (0.0821 L·atm/(mol·K) * 500 K)
ln(Q) ≈ 4.915
Taking the exponential of both sides to solve for Q:
Q ≈ e^(4.915)
Q ≈ 137.89
Now, let's calculate the partial pressure of hydrogen (P(H2)) using the given values of Q, P(HI), and P(I2):
Q = [HI]^2 / ([H2] * [I2])
137.89 = (10 atm)^2 / (P(H2) * 0.001 atm)
Rearranging the equation and solving for P(H2):
P(H2) = (10 atm)^2 / (137.89 * 0.001 atm)
P(H2) ≈ 775 atm
Therefore, the partial pressure of hydrogen (P(H2)) at this temperature should be approximately 775 atm in order to reduce the magnitude of ΔG for the reaction to zero.