Question

A reaction has an equilibrium constant of 8x103 at 298 K. At 716 K, the equilibrium constant is 0.77. Find AHºrxn for the reaction. Enter your answer numerically in kJ to 4 decimal places.

Answer 1

Step 1: Explanation

The Van't hoff equation allows us to calculate standard enthalpy change when two different temperature and two different rate constant is given.

we used the following given equation

ln ( k2/k1) = -ΔH0rxn / R(1/T2-1/T1)

where,

k represents the rate constant,

ΔH0rxn is the standard enthalpy chnage

R is the gas constant (8.3145 J/K mol),

and T is the temperature expressed in Kelvin

Step 2: Extract the data from question

Temperature (T1) = 298 K Rate constant (k1) = 8 × 103

Temperature (T2) = 716 K Rate constant (k2) = 0.77

Step 3: Calculation of standard enthalpy change (ΔH0rxn)

Given equation

=> ln ( k2/k1) = -ΔH0rxn/R(1/T2-1/T1)

on substituting the value in the equation

=> ln ( 0.77 / 8 × 103 ) =((-ΔH0rxn) / 8.314 J/ mol-1 K-1 )) (1/716 K -1/298K)

=> -9.248561585 =((-ΔH0rxn) / 8.314 J/ mol-1 K-1 )) ( -1.959056653 × 10-3 K-1)

=> ( -9.248561585 × 8.314 J/ mol-1 K-1 ) /  ( -1.959056653 × 10-3 K-1) = -ΔH0rxn

=> ΔH0rxn = -39249.8 J/mol ≈ -39.2498 kJ/mol

[ note: 1000 J = 1 kJ => -39249.8 J/mol × (1 kJ/ 1000 J) = -39.2498 kJ/mol ]

Hence, standard enthalpy change for the reaction ≈ -39.2498 kJ/mol