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____ [38]
3 years ago
9

C(graphite) + O2(g) → CO2(g) + 94.05 kcal C(diamond) + O2(g) → CO2(g) + 94.50 kcal What could you infer from the information giv

en above?
Chemistry
2 answers:
olga nikolaevna [1]3 years ago
7 0
Looking at both reactions, we can see that the combustion of carbon, graphite or diamond, leads to the formation of carbon dioxide and energy. Since energy is a product of the reaction, we known that this is an exothermic reaction. The value of the change in enthalpy, ΔH, will be negative.

A negative value of ΔH in each exothermic reaction suggests that the carbon dioxide product is at a lower energy than the reactants. And since more energy is released in the combustion of diamond compared to graphite, we know that diamond has a higher internal energy than graphite.
mr_godi [17]3 years ago
7 0

Answer: E 2 - E 1 is negative. and

Diamond has a higher enthalpy than graphite.

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<em><u>Answer and Explanation:</u></em>

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3 0
3 years ago
The standard molar heat of fusion of ice is 6020 j/mol. calculate q, w, and ∆e for melting 1.00 mol of ice at 0◦c and 1.00 atm p
zysi [14]

Answer :    q = 6020 J, w = -6020 J, Δe = 0

Solution : Given,

Molar heat of fusion of ice = 6020 J/mole

Number of moles = 1 mole

Pressure = 1 atm

Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.

The relation between heat and molar heat of fusion is,

q=\Delta H_{fusion}(\frac{Mass}{\text{ Molar mass}})  (in terms of mass)

or, q=\Delta H_{fusion}\times Moles     (in terms of moles)

Now we have to calculate the value of q.

q=6020J/mole\times 1Mole=6020J

When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.

So, the value of \Delta e=0

Now we have to calculate the value of w.

Formula used :    \Delta e=q+w

where, q is heat required, w is work done and \Delta e is internal energy.

Now put all the given values in above formula, we get

0=6020J+w

w = -6020 J

Therefore, q = 6020 J, w = -6020 J, Δe = 0

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