The standard enthalpy change for the following reaction is 873 kj at 298 k.
2 kcl(s) 2 k(s) + cl2(g) δh° = 873 kj
what is the standard enthalpy change for this reaction at 298 k?
k(s) + 1/2 cl2(g) kcl(

 - 436.5 kJ.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The given chemical reaction is,

Now we have to determine the value of for the following reaction i.e,

According to the Hess’s law, if we divide the reaction by half then the will also get halved and on reversing the reaction , the sign of enthlapy changes.

So, the value for the reaction will be:

Hence, the value of for the reaction is -436.5 kJ.

the answer is b. dilution