The equilibrium constant for the reaction of the weak base aniline is Kb = 4.3e-10.
C6H5NH2 (aq) + H2O (l) --> C6H5NH3+ (aq) + OH- (aq).
Calculate ΔG at 298 K when [C6H5NH2] = 0.250 M, [C6H5NH3+] = 1.0E-5 M, [OH-] = 0.000480 M
i used the equations dG=dG^o - RTlnQ and
dG^o= -RTlnK
and got the answer 97456.76003, but im told that answer is incorrect...what did i do wrong?
Isn't dG = dGo + RTlnQ?
To calculate ΔG at 298 K using the given concentrations, you need to make sure you have the correct values for ΔG° and the reaction quotient Q. It seems that you used the correct equations, but there might be an error in the values you have used or the units.
To find ΔG°, you can use the equation:
ΔG° = -RTln(K)
Given that the equilibrium constant Kb = 4.3e-10, you can plug in the values:
ΔG° = -(8.314 J/mol*K)(298 K)ln(4.3e-10)
ΔG° = -24451.239 J/mol
Now, let's calculate Q using the given concentrations:
Q = [C6H5NH3+][OH-] / [C6H5NH2]
Q = (1.0E-5)(0.000480) / 0.250
Q = 4.8E-12
Now, we can substitute the values of R (8.314 J/mol*K), T (298 K), ΔG° (-24451.239 J/mol), and Q (4.8E-12) into the equation:
ΔG = ΔG° - RTln(Q)
ΔG = -24451.239 - (8.314)(298)ln(4.8E-12)
Calculating this expression should yield the correct value for ΔG at 298 K.
To calculate ΔG at 298 K using the given concentrations and the equilibrium constant, let's follow these steps:
Step 1: Write down the expression for Q, the reaction quotient.
Q = ([C6H5NH3+][OH-]) / ([C6H5NH2][H2O])
Step 2: Calculate Q using the given concentrations.
Q = (1.0E-5 * 0.000480) / (0.250 * 1)
Step 3: Calculate ΔG^o, the standard Gibbs free energy change at 298 K, using the equation:
ΔG^o = -RT ln K
where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant.
Since the equilibrium constant (Kb) for the reaction of aniline is given as 4.3e-10, we can substitute the values into the equation:
ΔG^o = -(8.314 J/mol·K)(298 K) ln (4.3e-10)
Step 4: Calculate ΔG using the equation:
ΔG = ΔG^o - RT ln Q
Substitute the values of ΔG^o, R, T, and Q into the equation, and solve for ΔG:
ΔG = (ΔG^o - (8.314 J/mol·K)(298 K) ln Q)
Now, let's calculate the values:
ΔG^o = -(8.314 J/mol·K)(298 K) ln (4.3e-10)
= - 19245.272 J/mol
Q = (1.0E-5 * 0.000480) / (0.250 * 1)
= 1.92E-8
ΔG = (-19245.272 J/mol) - ((8.314 J/mol·K)(298 K) ln (1.92E-8))
≈ -19245.272 J/mol - 79613.487 J/mol
≈ -98858.76 J/mol
Therefore, the value of ΔG at 298 K, using the given concentrations and the equilibrium constant, is approximately -98858.76 J/mol.
If you obtained a different answer, please verify your calculations and double-check if you used the correct values for the equilibrium constant and concentrations.