Question

provide the value of the standard enthalpy change (δh°) for the following reaction: ch₃br + hcl → ch₃cl + hbr
bde values:
ch₃—br bond = 302 kj/mol
h—cl bond = 432 kj/mol
ch₃—cl bond = 350 kj/mol
h—br bond = 366 kj/mol
δh° = -114 kj/mol
δh° = 18 kj/mol
δh° = 114 kj/mol
δh° = -18 kj/mol

Explanation:

Step1: Identify bonds broken

Bonds broken are $CH_3 - Br$ and $H - Cl$.
$E_{broken}=302 + 432=734\ kJ/mol$

Step2: Identify bonds formed

Bonds formed are $CH_3 - Cl$ and $H - Br$.
$E_{formed}=350+366 = 716\ kJ/mol$

Step3: Calculate $\Delta H^{\circ}$

$\Delta H^{\circ}=E_{broken}-E_{formed}=734 - 716=18\ kJ/mol$

Answer:

$\Delta H^{\circ}=18\ kJ/mol$