# Enthalpy of A Reaction Using Hess' Law

This solution will discuss how to determine the enthalpy of a chemical reaction primarily using Hess's law. The use of the general formula for enthalpy will also be addressed. These problems will be solved using both Hess's law and the general formula for enthalpy. A step by step example and answers are included.

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Calculating Entropy using Hess's law

Find the enthalpy of the following reactions:

NOg + Â½ o2 NO2

Cgraphite C diamond

5C(s) +6H2(g) C5H12

1. There are 2 ways to solve these problems. One can solve them using Hess's law or one can solve them using the following formula:

âˆ‘ { (Î”H products) x (mol products)} - âˆ‘{ ( Î”H reactants) x ( mol reactants)}

I will solve them using hess's law and I will use the above formula to check my answers for some of the problems.

Formation of NO2

Add Î”H of each component reaction. Don't forget to reverse the reaction if necessary to obtain the desired product. If one reverses the reaction, one needs to assign the opposite sign to the Î”H value.

Write the equation for the formation of NO as follows:

Â½ N2(g) + Â½ o2(g) NO(g) Î” H = 90.29 kj

We must reverse this reaction because we want a reaction in which NO is the reactant. We can do this because according to Hess's law, it does not matter which way a reaction proceeded. So we have the following equation:

NO(g) Â½ N2(g) + Â½ o2 Î”H=-90.29kj

Â½ N2(g) + o2(g) NO2(g) Î”H=33.2kj

NO(G) + Â½ o2(g)* NO2 Î”H=-57.1kj

*o2 partially cancels, If you cancel out Â½ o2. That leaves you with Â½ o2.

You can check the answer using the general formula for enthalpy, as mentioned previously. If one substitutes Î”H values it is as follows:

33.2 kj/mol - 90.29kj/mol = -57.1 kj

Since o2 is in its ground state or natural state, it's Î”H value is 0.

2. Cgraphite C diamond

Cgraphite + o2 co2(g) Î”H = -394 kj

Cdiamond + o2 co2(g) Î”H= -396 kj

Keep the 1st equation as is but reverse the second because cdiamondhas to be a product.

C graphite + O2 CO2 Î”H = -394 kj

CO2 Cdiamond + O2 Î”H =+ 396kj

Cgraphite C diamond Î”H = 2 kj

Check with formula

-394 kj + 396 kj = 2 kj

3. Formation of Pentane

5C(s) +6 H2(g) C5H12

For this reaction one needs to use the combustion reactions of each of the reactants as follows:

C(s) + O2(g) CO2(g) Î”H = -393.5 kj

H2(g) + Â½ O2 H2O(l) Î”H = -285.8 kj

C5H12(g) + 8O2(g) 5CO2(g) + 6 H2O(l) Î”H=-3535.6kj

By combining equations we can solve this as follows:

5C(s) + 5O2(g) 5 CO2(g) Î”H=5(-393.5)kj

6 H2 + 3O2 6H2O Î”H=6(-285.8)kj

5CO2 +6H2O C5H12 + 8O2 Î”H=3535.6 kj

5 C + 6H2 C5H12 Î”H=146.7 kj

Using the general formula for enthalpy we can compute the answer as follows:

-3535.6-[5(-393.5)+6(-285.8)]

-3535.6-{(-1967.5 +(-1714.8)}

-3535.6-(-3682.3)

-3535.6+3682.3=146.7

Modern Chemistry,Holt, Rinehart, Winston 2006 Austin TX, Raymond E. Davis et al.

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