At 25^° C, the enthalpy of the following processes are given : H_2(g)+O_2(g) -> 2 OH(g) Delta H^°=78 kJ mol^-1 H_2(g)+(1 / 2)O_2(g) -> H_2O(g) Delta H^°=-242 kJ mol^-1 H_2(g) -> 2H(g) Delta H^°=436 kJ mol^-1 (1 / 2) O_2(g) -> O(g) Delta H^°=249 kJ mol^-1 What would be the value of X for the following reaction ? _____________ (Nearest integer) H_2O(g) -> H(g)+OH(g) Delta H^°=X kJ mol^-1

frac (i) + (iii) 2- (ii) gives desired reaction Delta H_ r=(436+78 / 2)-(-242) \ =(436+78 / 2)+242=499

At 25 C the enthalpy of the following processes are giv... | (Chemistry) | DivineJEE

Exam

Chemistry

At $25^{\circ} \mathrm{C}$ , the enthalpy of the following processes are given :

$\mathrm{H_2(g)+O_2(g)}$	$\to$	$2\mathrm{OH(g)}$	$\mathrm{\Delta H^\circ=78~kJ~mol^{-1}}$
$\mathrm{H_2(g)+\frac{1}{2}O_2(g)}$	$\to$	$\mathrm{H_2O(g)}$	$\mathrm{\Delta H^\circ=-242~kJ~mol^{-1}}$
$\mathrm{H_2(g)}$	$\to$	$\mathrm{2H(g)}$	$\mathrm{\Delta H^\circ=436~kJ~mol^{-1}}$
$\frac{1}{2}\mathrm{O_2(g)}$	$\to$	$\mathrm{O(g)}$	$\mathrm{\Delta H^\circ=249~kJ~mol^{-1}}$