Question

Given, H2(g) + Br2(g) $\longrightarrow$ 2HBr(g), $\Delta$H01 and standard enthalpy of condensation of bromine is $\Delta$H02, standard enthalpy of formation of HBr at 250C is

• A. $\Delta$H01 / 2
• B. $\Delta$H01 / 2 + $\Delta$H02
• C. $\Delta$H01 / 2 - $\Delta$H02
• D. ($\Delta$H01-$\Delta$H02) / 2

Answer 1

Answer: (D)

($\Delta$H01-$\Delta$H02) / 2

Answer 2

H2(g) + Br2(g) $\longrightarrow$ 2HBr(g)$\Delta$H = $\Delta$H°1 ...(i)

Br2(g) $\longrightarrow$ Br2($\mathcal{l}$) $\Delta$H = $\Delta$H°2 ...(ii)

[eq1 – eq2 ]

H2(g) + Br2(l) $\longrightarrow$2HBr(g)$\Delta$H = $\Delta$H°1 –$\Delta$H°2

Required equation, $\frac{1}{2}$H2(g) + $\frac{1}{2}$Br2(l) $\longrightarrow$ HBr(g) $\Delta$H =$\left\lbrack \frac{\Delta H_{1}^{0} - \Delta H_{2}^{0}}{2} \right\rbrack$