Question

What will be the enthalpy change of conversion of graphite into diamond?
Given
C graphite, ∆combH = -391.25 kJ;
C diamond, AcombH = - 393.12 kJ

C(graphite)---------> C(diamond)

( please also write when to do R-P and P-R)

R= reactant⟩
P= Product⟩

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Answer 1

Answer:

Given Cgraphite,△combH = -391.25 kJ; Cdiamond,△combH = -393.12 kJ.

Explanation:

hope it's helpful

❤️sam❤️

Answer 2

Answer:

1.87 kJ

Explanation:

For doing thermodynamics problems always observe which Heat of reaction is given. Don't just go with the formula, because it is confusing

We will get confusion where to take R-P or P-R

But, the basic formula is

$\Delta H_{reaction } = \sum Bond \ energies\ of\ reactants\ - \sum Bond\ energies \ of\ products$

But, this formula is applicable only when all components are in gaseous state.

So, to avoid all these confusions, see the Heat given

Here given is 'comb' which means Combustion

So, first write the combustion reaction and then solve the problem

The reactions are,

$C_{graphite} + O_2 \longrightarrow CO_2 \ ; \Delta H_1 = -391.25\ kJ \ \ ...[1]\\C_{diamond} + O_2 \longrightarrow CO_2 \ ; \Delta H_2 = -393.12 \ kJ \ \ ... [2]$

So, these are the combustion reactions

Now, modifications;

In the question C(diamond) is on the right side, but in combustion reaction it is on the left side. So, we must flip the second reaction

Now, it turns into,

$CO_2 \longrightarrow C_{diamond} + O_2 \ ; \Delta H_3 = 393.12\ kJ \ \ ...[3]$

As the reaction flipped, ∆H sign changes

Now, Just directly add equation [1] and [3], which is the basic method of solving by Hess law

$C_{graphite} + O_2 \longrightarrow CO_2 \ ; \Delta H_1 = -391.25\ kJ \\CO_2 \longrightarrow C_{diamond} + O_2 \ ; \Delta H_3 = 393.12\ kJ \\\text{ -----------------------------------------------------------}\\C_{graphite} \longrightarrow C_{diamond} \ ; \ \Delta H = \Delta H_1 + \Delta H _3$

So, ΔH = ΔH₁+ΔH₃

=> ΔH = -391.25 + 393.12

=> ΔH = 1.87 kJ

[If same components are on right and left side of reaction, they cancel out. CO₂ and O₂ are on both sides, so they are canceled out]

Try doing some problems in this method, this is the easiest method which I made myself through practice. Even other heats like bond dissociation, formation, atomization can also be done by this method. Try few problems. You will get the right answers. I can't post answers for some days as my exams are nearby. But if possible I will answer your questions with some tricks and modifications

Hope it helps!