H_2 + Cl_2 longrightarrow 2HCl ; Delta H = -x | vedclass

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(A) The enthalpy of formation $(\Delta H_f)$ of a compound is defined as the enthalpy change when $1 \ mole$ of the compound is formed from its consti

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$-\frac{x}{2} \ kJ$

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(A) The enthalpy of formation $(\Delta H_f)$ of a compound is defined as the enthalpy change when $1 \ mole$ of the compound is formed from its constituent elements in their standard states. The reaction for the formation of $HCl$ is: $\frac{1}{2}H_2(g) + \frac{1}{2}Cl_2(g) \longrightarrow HCl(g)$. From the first given equation: $H_2(g) + Cl_2(g) \longrightarrow 2HCl(g)$ ; $\Delta H = -x \ kJ$. Dividing the entire equation and the enthalpy change by $2$,we get: $\frac{1}{2}H_2(g) + \frac{1}{2}Cl_2(g) \longrightarrow HCl(g)$ ; $\Delta H = -\frac{x}{2} \ kJ$. Thus,the $\Delta H_f$ of $HCl$ is $-\frac{x}{2} \ kJ$.

$H_2 + Cl_2 \longrightarrow 2HCl$ ; $\Delta H = -x \ kJ$
$NaCl + H_2SO_4 \longrightarrow NaHSO_4 + HCl$ ; $\Delta H = -y \ kJ$
$2H_2O + 2Cl_2 \longrightarrow 4HCl + O_2$ ; $\Delta H = -z \ kJ$
From the above equations,the value of $\Delta H_f$ of $HCl$ is:

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$H_2 + Cl_2 \longrightarrow 2HCl$ ; $\Delta H = -x \ kJ$ $NaCl + H_2SO_4 \longrightarrow NaHSO_4 + HCl$ ; $\Delta H = -y \ kJ$ $2H_2O + 2Cl_2 \longrightarrow 4HCl + O_2$ ; $\Delta H = -z \ kJ$ From the above equations,the value of $\Delta H_f$ of $HCl$ is: