Chemical Equilibrium Question 7
Question 7 - 01 February - Shift 1
At $25^{\circ} C$, the enthalpy of the following processes are given:
$H_2(g)+O_2(g) \to 2 OH(g) \Delta H^{0}=78 kJ mol^{-1}$
$H_2(g)+1 / 2 O_2(g) \to H_2 O(g) \Delta H^{o}=-242 kJ mol^{-1}$
$H_2(g) \to 2 H(g) \Delta H^{0}=436 kJ mol^{-1}$
$1 / 2 O_2(g) \to O(g) \Delta H^{0}=249 kJ mol^{-1}$
What would be the value of $X$ for the following reaction? (Nearest integer)
$H_2 O(g) \to H(g)+OH(g) \Delta H^{0}=X kJ mol^{-1}$
Show Answer
Answer: (499)
Solution:
Formula: Calculation of enthalpy of a reaction
$2 H_2 O(g) \to 2 H_2(g)+O_2(g) \quad+(242 \times 2) kJ mol^{-1}$
$H_2(g)+O_2(g) \to 2 OH \quad+78 kJ mol^{-1}$
$H_2(g) \to 2 H \quad+436 kJ mol^{-1}$
$2 H_2 O \to 2 H+2 OH \quad+998 kJ mol^{-1}$
$H_2 O \to H+OH \quad 998 \times \frac{1}{2}=+499 kJ mol^{-1}$
