Question

The standard enthalpy change for the following reaction is 220 kJ at 298 K.

CuCl₂(s) -->Cu(s) + Cl₂(g)   ΔH° = 220 kJ

What is the standard enthalpy change for this reaction at 298 K?

Cu(s) + Cl2(g) --> CuCl2(s)

Answer in kJ

Answer 1

The standard enthalpy change for the reaction is 220 kJ at 298 K.

CuCl₂(s) -->Cu(s) + Cl₂(g)   ΔH° = 220 kJ

given the reaction

Cu(s) + Cl₂(g) --> CuCl₂(s)

the above reaction is reverse of the standard reaction

so delta H will be same in magnitude and but the sign will be opposite

so

Cu(s) + Cl₂(g) --> CuCl₂(s)       ΔH° =    - 220 kJ (it is -ve value )