Question

$C{l_{(g)}} + {e^ - } \to Cl_{(g)}^ -$, ${{\text{E}}_A}$ = - 348 KJ/mol at 0K, then ΔH for same process at 300 K (in KJ) [R = 8 J/mol/K]
[Hint : ΔH = ${{\text{E}}_A}$ - $\dfrac{5}{2}$RT]

Answer 1

The ΔH represents the change in enthalpy of a system. It can be calculated by the formula given in the hint part of the question as -
ΔH = ${{\text{E}}_A}$ - $\dfrac{5}{2}$RT
Where ΔH = change in enthalpy
${{\text{E}}_A}$= activation energy of the system (given)
R = gas constant
T = temperature of the system

Complete step by step answer:
This is the most simple question. Everything is given in question. We should just observe and then it is easy to solve. Let us first write the things given to us in question.
Given :
$C{l_{(g)}} + {e^ - } \to Cl_{(g)}^ -$
${{\text{E}}_A}$ = - 348 KJ/mol
Temperature = 0 K
R = gas constant = 8 J/mol/K
Further, in the hint we have been given this formula -
ΔH = ${{\text{E}}_A}$ - $\dfrac{5}{2}$RT
To find :
ΔH for same process at 300 K (in KJ)
So, as we have been given the formula -
ΔH = ${{\text{E}}_A}$ - $\dfrac{5}{2}$RT
Let us put all the values in this, the temperature will be now 300 K
So, ΔH = - 348 - $\dfrac{5}{2}$ $\times$8$\times$300$\times$ ${10^{ - 3}}$
ΔH = - 348 - 6
ΔH = - 354 KJ

So, this is our answer.

Note: It must be noted that 1 KJ is equal to 1000 J. So, to convert J into KJ; we need to divide by 1000. We have initially converted the J into KJ because the activation energy given to us was in KJ and the value of R was in J. So, to subtract them both should be in the same units.