The standard enthalpy of formation of H 2(g) and Cl 2(g) and HCl (g) are 218 kJ / mol, 121.68 kJ / mol and 92.31 kJ / mol respectively. Calculate standard enthalpy change in k J for (1/2) H2(g), (1/2) C l2(g) arrow H C l(g)

Question

The standard enthalpy of formation of H 2(g) and Cl 2(g) and HCl (g) are 218 kJ / mol, 121.68 kJ / mol and 92.31 kJ / mol respectively. Calculate standard enthalpy change in k J for (1/2) H2(g), (1/2) C l2(g) arrow H C l(g) (A) + 431.99 (B) - 246.37 (C) - 431.99 (D) + 247.37

Tardigrade Admin · Accepted Answer

For reaction, (1/2) H 2(g)+(1/2) Cl 2(g) arrow HCl (g) Δ H =Δ H f( HCl ) -[(1/2) × Δ H f( H 2)+(1/2) × Δ H f( Cl 2)] =-92.31-[(1/2) ×(218)+(1/2) ×(121.68)] =-262.15 kJ

Q. The standard enthalpy of formation of $H_{2} (g)$ and $C l_{2} (g)$ and $H C l_{(g)}$ are $218 k J / m o l, 121.68 k J / m o l$ and $92.31 k J / m o l$ respectively. Calculate standard enthalpy change in $k J$ for $\frac{1}{2} H_{2} (g), \frac{1}{2} C l_{2} (g) \to H C l_{(g)}$

+ 431.99

- 246.37

- 431.99

+ 247.37

-262.15

For reaction, $\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H Cl (g)$
$Δ H = Δ H_{f} (H Cl)$
$- [\frac{1}{2} \times Δ H_{f} (H_{2}) + \frac{1}{2} \times Δ H_{f} (C l_{2})]$
$= - 92.31 - [\frac{1}{2} \times (218) + \frac{1}{2} \times (121.68)]$
$= - 262.15 k J$

Q. The standard enthalpy of formation of H2​(g) and Cl2​(g) and HCl(g)​ are 218kJ/mol,121.68kJ/mol and 92.31kJ/mol respectively. Calculate standard enthalpy change in kJ for 21​H2​(g),21​Cl2​(g)→HCl(g)​