AB,A_2,and B_2 are diatomic molecules. Enthal | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(400) Let the dissociation enthalpy (bond energy) of $A_2$ be $X$. Given the ratio $1:1:0.5$ for $AB:A_2:B_2$,the bond energy of $AB$ is $X$ and the b

Vedclass · Accepted Answer

Vedclass · Answer

(400) Let the dissociation enthalpy (bond energy) of $A_2$ be $X$.
Given the ratio $1:1:0.5$ for $AB:A_2:B_2$,the bond energy of $AB$ is $X$ and the bond energy of $B_2$ is $0.5X$.
The enthalpy of formation reaction is: $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB ; \Delta_f H = -100 \ kJ \ mol^{-1}$.
Using the bond energy formula: $\Delta_f H = \Sigma BE_{\text{reactants}} - \Sigma BE_{\text{products}}$.
$-100 = (\frac{1}{2} \times BE_{A_2} + \frac{1}{2} \times BE_{B_2}) - BE_{AB}$.
$-100 = (\frac{1}{2} X + \frac{1}{2} \times 0.5X) - X$.
$-100 = (0.5X + 0.25X) - X$.
$-100 = 0.75X - X$.
$-100 = -0.25X$.
$X = \frac{100}{0.25} = 400 \ kJ \ mol^{-1}$.

$AB$,$A_2$,and $B_2$ are diatomic molecules. Enthalpies of dissociation of $AB$,$A_2$,and $B_2$ are in the ratio of $1:1:0.5$. Enthalpy of formation of $AB$,$\Delta_f H = -100 \ kJ \ mol^{-1}$. Find the dissociation enthalpy of $A_2$?
Reaction : $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB$

Similar Questions

The enthalpy of formation $(\Delta H_f)$ of methanol,formaldehyde and water are $-239, -116$ and $-286 \ kJ \ mol^{-1}$ respectively. The enthalpy change for the oxidation of methanol to formaldehyde and water in $kJ$ is

Which thermochemical reaction is correct?

In an exothermic reaction,$\Delta H$ is

If $C + O_2 \to CO_2 + 94.2 \ kcal$,$H_2 + \frac{1}{2} O_2 \to H_2O + 68.3 \ kcal$,and $CH_4 + 2O_2 \to CO_2 + 2H_2O + 210.8 \ kcal$,then the possible heat of formation of methane will be $...... \ kcal$.

The enthalpy of reaction for the reaction: $2 H_{2(g)} + O_{2(g)} \to 2 H_{2}O_{(l)}$ is $\Delta_{r}H^{\theta} = -572 \ kJ \ mol^{-1}$. What will be the standard enthalpy of formation of $H_{2}O_{(l)}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$AB$,$A_2$,and $B_2$ are diatomic molecules. Enthalpies of dissociation of $AB$,$A_2$,and $B_2$ are in the ratio of $1:1:0.5$. Enthalpy of formation of $AB$,$\Delta_f H = -100 \ kJ \ mol^{-1}$. Find the dissociation enthalpy of $A_2$?Reaction : $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB$