When K is measured for a reaction, the values obtained are 3.1 at 25°C and 7.42 at 35°C. Using the equation lnK=-(∆H°/R)(1/T)+(∆S°/R) (in the form y=mx+b) calculate ∆H° and ∆S° for the reaction PbCl2(s) --> Pb2+(aq) + Cl-(aq)
To find the values of ∆H° (enthalpy change) and ∆S° (entropy change) for the reaction, we can use the relationship between the natural logarithm of the equilibrium constant (lnK) and the temperature (T) according to the equation:
lnK = -(∆H°/R)(1/T) + (∆S°/R)
where:
lnK is the natural logarithm of the equilibrium constant (K),
∆H° is the standard enthalpy change,
∆S° is the standard entropy change,
R is the ideal gas constant (8.314 J/(mol·K)), and
T is the temperature in Kelvin (K).
We are given two sets of data at different temperatures (25°C and 35°C) with their corresponding values of K. Let's convert the temperatures to Kelvin:
25°C = 25 + 273.15 = 298.15 K
35°C = 35 + 273.15 = 308.15 K
Now, let's rewrite the equation in the form y = mx + b, where y = lnK, x = 1/T, m = -∆H°/R, and b = ∆S°/R.
For the first set of data at 25°C:
lnK₁ = -(∆H°/R)(1/T₁) + (∆S°/R)
ln(3.1) = -(∆H°/R)(1/298.15) + (∆S°/R) ----(1)
For the second set of data at 35°C:
lnK₂ = -(∆H°/R)(1/T₂) + (∆S°/R)
ln(7.42) = -(∆H°/R)(1/308.15) + (∆S°/R) ----(2)
Now, we have two equations (1) and (2) with two unknowns (∆H° and ∆S°). To solve for these unknowns, let's subtract equation (2) from equation (1):
ln(3.1) - ln(7.42) = [(∆H°/R)(1/298.15) - (∆H°/R)(1/308.15)] + [∆S°/R - ∆S°/R]
ln(3.1/7.42) = (∆H°/R)[(1/298.15) - (1/308.15)]
Using the property ln(a/b) = ln(a) - ln(b), we simplify it further:
ln(3.1/7.42) = (∆H°/R)[(308.15 - 298.15)/(298.15 * 308.15)]
Now, let's solve for ∆H°/R:
(∆H°/R) = [ln(3.1/7.42)] / [(308.15 - 298.15)/(298.15 * 308.15)]
Evaluate the right side of the equation:
(∆H°/R) = [ln(3.1/7.42)] / [(10.0)/(298.15 * 308.15)]
Now, substitute the value of R (the ideal gas constant) as 8.314 J/(mol·K) and evaluate (∆H°/R):
(∆H°/8.314) = [ln(3.1/7.42)] / [(10.0)/(298.15 * 308.15)]
(∆H°/8.314) = ln(3.1/7.42) / (10.0 * 298.15 * 308.15)
Finally, multiply both sides by 8.314 to solve for ∆H°:
∆H° = [ln(3.1/7.42) / (10.0 * 298.15 * 308.15)] * 8.314
Calculate this expression using a calculator to find the value of ∆H°.
Once you have ∆H°, you can rearrange the equation to solve for ∆S°:
lnK = -(∆H°/R)(1/T) + (∆S°/R)
lnK - (-∆H°/R)(1/T) = (∆S°/R)
(∆S°/R) = lnK + (∆H°/R)(1/T)
Now, substitute the values of lnK, ∆H°, R, and T to calculate ∆S°.