$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$
- A$-3.8 \ kJ/mole$
- B$-1.9 \ kJ/mole$
- C$3.8 \ kJ/mole$
- D$1.9 \ kJ/mole$
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Similar Questions
The heat of neutralization of $HCl$ by $NaOH$ is $-57.3 \, kJ/mol$. If the heat of neutralization of $HCN$ by $NaOH$ is $-12.1 \, kJ/mol$,then the enthalpy of dissociation of $HCN$ is $...... \, kJ$.
Easy
View SolutionThe bond dissociation energies of $X_2$,$Y_2$ and $XY$ are in the ratio of $1 : 0.5 : 1$. $\Delta H$ for the formation of $XY$ is $-200 \ kJ \ mol^{-1}$. The bond dissociation energy of $X_2$ will be $...... \ kJ \ mol^{-1}$.
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The heat of formation of $CO_2$ is ..... $kJ$.
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The heat of formation of $CO_2$ is ..... $kJ$.
Medium
View SolutionThe bond energies of $C-C$,$C=C$,$H-H$,and $C-H$ bonds are $350$,$600$,$400$,and $410 \ kJ \cdot mol^{-1}$ respectively. The heat of hydrogenation of ethylene $(C_2H_4)$ is ... $kJ \cdot mol^{-1}$.
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View SolutionThe bond dissociation enthalpies of $H_2$,$Cl_2$,and $HCl$ are $434$,$242$,and $431 \ kJ \ mol^{-1}$ respectively. What is the enthalpy of formation of $HCl$ in $kJ \ mol^{-1}$?
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