If Kc = 0.480 at 40.°C and Kc = 0.630 at 90.°C, what is ΔH° for the reaction?

X<---->Y

Use the van't Hoff equation.

To calculate the standard enthalpy change (ΔH°) for the given reaction, we can use the Van 't Hoff equation:

ln(K2/K1) = ΔH°/R * (1/T1 - 1/T2)

Where:
K1 and K2 are the equilibrium constants at temperatures T1 and T2 respectively,
ΔH° is the standard enthalpy change,
R is the gas constant (8.314 J/mol·K), and
T1 and T2 are the temperatures in Kelvin.

In this case, we have:
K1 = 0.480 at 40.°C (313 K),
K2 = 0.630 at 90.°C (363 K),
R = 8.314 J/mol·K.

Let's plug in the values and solve for ΔH°:

ln(0.630/0.480) = ΔH°/8.314 * (1/313 - 1/363)

ln(1.313) = ΔH°/8.314 * (0.003194 - 0.002753)

We can simplify the equation further:

0.2748 = ΔH°/8.314 * 0.000441

Now solve for ΔH° by rearranging the equation:

ΔH° = 0.2748 * 8.314 / 0.000441

ΔH° = 5173.45 J/mol

Therefore, the standard enthalpy change (ΔH°) for the given reaction is +5173.45 J/mol.

To determine the value of ΔH° for the reaction X ⇌ Y, we can use the equation ΔH° = -R * T * ln(Kc), where R is the gas constant, T is the temperature in Kelvin, and ln is the natural logarithm.

First, we need to convert the given temperatures from degrees Celsius to Kelvin. The temperature of 40.°C would be 313.15 K (40 + 273.15), and the temperature of 90.°C would be 363.15 K (90 + 273.15).

Next, we can use the equation ΔH° = -R * T * ln(Kc) to calculate ΔH° at both temperatures:

At 40.°C (313.15 K):
ΔH° = -R * T * ln(Kc)
ΔH° = -R * 313.15 K * ln(0.480)

At 90.°C (363.15 K):
ΔH° = -R * T * ln(Kc)
ΔH° = -R * 363.15 K * ln(0.630)

To get the final value for ΔH°, we need to subtract the two ΔH° values calculated at different temperatures:

ΔH° = (ΔH° at 90.°C) - (ΔH° at 40.°C)
ΔH° = -R * 363.15 K * ln(0.630) - (-R * 313.15 K * ln(0.480))

Where R is the gas constant (8.314 J/(mol·K)).

Note: Make sure to use consistent units throughout the calculation.