Question

For the following reaction,
$C(diamond,s)+O_2\left(g\right)\to C{O}_2\left(g\right);△H=-94.3kcal$
$C(graphite,s)+O_2\left(g\right)\to C{O}_2\left(g\right);△H=-97.6kcal$
The heat required to change 1 g of $C(diamond,s)\to C(graphite,s)$ is:

• A. $3.3kcal$
• B. $1.65kcal$
• C. $1.29kcal$
• D. $0.275kcal$

Answer 1

The correct option is D $0.275kcal$

$C(diamond,s)+O_2\left(g\right)\to C{O}_2\left(g\right);△H=-94.3kcal$
$C(graphite,s)+O_2\left(g\right)\to C{O}_2\left(g\right);△H=-97.6kcal$
using hess law,
$C(diamond,s)\to C(Graphite,s)△H=3.3kcal$

Heat required to convert 12 g diamond to graphite = 3.3
$\therefore$ Heat required to convert 1 g diamond to graphite $=\frac{3.3}{12}=0.275kcal$




Theory :

Lavoisier - Laplace Law:
If the direction of a chemical process is reversed then magnitude of enthalpy change remains the same but the sign is reversed.

This is because the enthalpy is a state function that means it is independent of path followed. So when we reverse the initial and final state its magnitude remains the same but the sign is reversed.

Reaction - 1 :
$A\to B$ ${\Delta }_{rx{n}_1}H$

Reaction - 2:
$B\to A$ ${\Delta }_{rx{n}_2}H$

${\Delta }_{rx{n}_1}H=-{\Delta }_{rx{n}_2}H$

For example

$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right)$ $\Delta H_1=-286kJ/mol$

$H_{2}O\left(l\right)\to H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$ $\Delta H_2=286kJ/mol$

Since here direction of the chemical process is reversed,

$\Delta H_1=-\Delta H_2$

Scaling of the reaction ${\Delta }_{rxn}H$
$A\to B$ x

$2A\to 2B$ 2x

$\frac{1}{2}A\to \frac{1}{2}B$ 1/2x

$nA\to nB$ $nx$