For the allotropic change represented by the | vedclass

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

Question

(D) The given reaction is $C(	ext{diamond}) 	o C(	ext{graphite})$ with $\Delta H = -1.89 \ kJ$. This means diamond is at a higher energy level than

Vedclass · Accepted Answer

More than in the second case by $0.945 \ kJ$

Vedclass · Answer

(D) The given reaction is $C(	ext{diamond}) 	o C(	ext{graphite})$ with $\Delta H = -1.89 \ kJ$. This means diamond is at a higher energy level than graphite by $1.89 \ kJ/mol$. When $1 \ mol$ $(12 \ g)$ of diamond burns,it releases $1.89 \ kJ$ more heat than $1 \ mol$ of graphite. For $6 \ g$ of diamond,the amount of substance is $6/12 = 0.5 \ mol$. Therefore,the extra heat liberated compared to graphite is $0.5 	imes 1.89 \ kJ = 0.945 \ kJ$. Thus,the heat liberated in the first case (diamond) is more than in the second case (graphite) by $0.945 \ kJ$.

For the allotropic change represented by the equation $C(\text{diamond}) \to C(\text{graphite})$,the enthalpy change is $\Delta H = -1.89 \ kJ$. If $6 \ g$ of diamond and $6 \ g$ of graphite are separately burnt to yield carbon dioxide,the heat liberated in the first case is:

Similar Questions

$A_2 + B_2 \rightarrow 2AB$; $\Delta H_{r}^0 = -400\,kJ\,mol^{-1}$. $AB$,$A_2$ and $B_2$ are diatomic molecules. If the bond enthalpies of $A_2$,$B_2$ and $AB$ are in the ratio $1:0.5:1$,then the bond enthalpy of $A_2$ is $......\,kJ\,mol^{-1}$ (Nearest integer).

$AB$,$A_2$,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 : $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB$

The heat of neutralization of a strong acid and a weak base is:

Heat produced in calories by the combustion of one gram of carbon is called

The bond dissociation energies of $XY$,$X_2$,and $Y_2$ (all diatomic molecules) are in the ratio $1 : 1 : 0.5$. If the enthalpy of formation of $XY$ is $\Delta_fH = -200 \ kJ \ mol^{-1}$,find the bond dissociation energy of $X_2$ in $kJ \ mol^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module