If the combustion of $1 \ g$ of graphite produces $20.7 \ kJ$ of heat,what will be the molar enthalpy change? Give the significance of the sign also.
(N/A) The enthalpy of combustion of $1 \ g$ of graphite is $20.7 \ kJ$.
Since heat is released,the enthalpy change for $1 \ g$ is $-20.7 \ kJ \ g^{-1}$.
The molar mass of graphite $(C)$ is $12 \ g \ mol^{-1}$.
The molar enthalpy change $(\Delta H)$ is calculated as:
$\Delta H = -20.7 \ kJ \ g^{-1} \times 12 \ g \ mol^{-1} = -248.4 \ kJ \ mol^{-1} = -2.48 \times 10^{2} \ kJ \ mol^{-1}$.
The negative sign indicates that the combustion reaction is exothermic,meaning heat is released to the surroundings.
Since heat is released,the enthalpy change for $1 \ g$ is $-20.7 \ kJ \ g^{-1}$.
The molar mass of graphite $(C)$ is $12 \ g \ mol^{-1}$.
The molar enthalpy change $(\Delta H)$ is calculated as:
$\Delta H = -20.7 \ kJ \ g^{-1} \times 12 \ g \ mol^{-1} = -248.4 \ kJ \ mol^{-1} = -2.48 \times 10^{2} \ kJ \ mol^{-1}$.
The negative sign indicates that the combustion reaction is exothermic,meaning heat is released to the surroundings.
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