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How many kJ of energy will be released when 4.72g of carbon react with excess oxygen to produce carbon dioxide (delta H is -393.5 kJ)

How many kJ of energy will be released when 4.72g of carbon react with excess oxygen-example-1
User Iamrajshah
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1 Answer

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Answer:

155 kJ of energy will be released.

Step-by-step explanation:

The
\rm \Delta H\textdegree (
\rm \Delta H \textdegree_\text{rxn} in some textbooks) here stands for standard enthalpy change per mole reaction. To find the amount of energy released in this reaction, start by finding the number of moles of this reaction that will take place.

How many moles of atoms in 4.72 grams of carbon?

Relative atomic mass data from a modern periodic table:

  • C: 12.01.


\displaystyle n = (m)/(M) = \rm (4.72\;g)/(12.01\; g\cdot mol^(-1)) = 0.393006\; mol.

The coefficient of carbon in the equation is one. In other words, each mole of the reaction will consume one mole of carbon. Oxygen is in excess. As a result,
\rm 0.393006\; mol of carbon will support
\rm 0.393006\; mol of the reaction.

How much energy will be released?

The
\rm \Delta H\textdegree{} value here is negative. But don't panic.
\rm \Delta H\textdegree{} is the same as the chemical potential energy of the reactants minus the products in one mole of the reaction.
\rm \Delta H\textdegree{} = -393.5\;kJ means that the chemical potential energy drops by
\rm 393.5\; kJ during each mole of the reaction (with the coefficients as-is.) Those energy difference will be released as heat. In other words, one mole of the reaction will release
\rm 393.5\;kJ of energy.

The 4.72 grams of carbon will support
\rm 0.393006\; mol of this reaction. How much heat will that
\rm 0.393006\; mol of reaction release?


Q = n \cdot (-\Delta \text{H}\textdegree{}) = \rm 0.393006* 393.5 = 155\;kJ.

As a side note, the mass of carbon 4.72 grams is the least significant data in this question. There are three significant figures in this value. As a result, keep more than three significant figures in calculations but round the final result to three significant figures.

User Katina
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