The equation frac{1}{2} H_2 + frac{1}{2} Cl_2 | vedclass

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(B) The given equation is $\frac{1}{2} H_2 + \frac{1}{2} Cl_2 	o HCl$ with $\Delta H_{298} = -22.060 \ kcal$.Since the enthalpy change $\Delta H$ is

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The heat evolved when one gram molecule of $HCl$ is formed from its elements at $298 \ K$ is $22.060 \ kcal$.

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(B) The given equation is $\frac{1}{2} H_2 + \frac{1}{2} Cl_2 	o HCl$ with $\Delta H_{298} = -22.060 \ kcal$.Since the enthalpy change $\Delta H$ is negative,the reaction is exothermic,meaning heat is released (evolved).The reaction represents the formation of $1 \ 	ext{mole}$ (one gram molecule) of $HCl$ from its constituent elements in their standard states at $298 \ K$.Therefore,the heat evolved during the formation of $1 \ 	ext{mole}$ of $HCl$ is $22.060 \ kcal$.

The equation $\frac{1}{2} H_2 + \frac{1}{2} Cl_2 \to HCl$ $(\Delta H_{298} = -22.060 \ kcal)$ means:

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