If the standard molar enthalpy change for combustion of graphite powder is -2.48 x 10^2 kJ mol^-1, t

Q: Why do we use 12 as the molar mass for graphite?

Graphite is a naturally occurring allotrope of carbon. It is composed entirely of carbon atoms. Therefore, its molar mass is simply the atomic mass of carbon, which is $12 \text{ g mol}^{-1}$.

Q: Why is the enthalpy change given as a negative value?

Combustion is an exothermic process, meaning the system loses energy (releases heat) to the surroundings. In thermodynamics, energy leaving the system is denoted with a negative sign. However, when asked for 'heat generated', we usually provide the magnitude (positive value).

Q: How do we handle the $10^2$ term in the calculation?

It is just scientific notation. $2.48 \times 10^2$ is mathematically equivalent to $2.48 \times 100$, which equals $248$. It is often easier to convert it to a standard number before dividing.

Q: What if the question asked for heat absorbed?

Combustion reactions always release heat; they never absorb it. Therefore, the heat of combustion is always an exothermic quantity. If a question asks for heat absorbed during combustion, it is a trick question, or the value would technically be negative.

Q: Why do we round 20.66 to 21 and not 20?

When rounding to the nearest integer, we look at the first decimal digit. If it is 5 or greater, we round up to the next whole number. Since 6 is greater than 5, $20.66$ rounds up to $21$.

Question

If the standard molar enthalpy change for combustion of graphite powder is $-2.48 	imes 10^2 	ext{ kJ mol}^{-1}$, the amount of heat generated on combustion of $1 	ext{ g}$ of graphite powder is ...... kJ. (Nearest integer)

Sigma Percentile · Accepted Answer

### The Burning Question: Calculating Heat of Combustion

Thermochemistry often feels like an abstract accounting of invisible energy, but it is deeply rooted in physical reality. When we burn a substance, the chemical bonds break and reform, releasing energy into the surroundings. This problem asks us to quantify that exact energy release for a tiny, specific amount of graphite.

Let's break down the problem. We are given the **standard molar enthalpy of combustion** for graphite powder, which is $\Delta_c H^\circ = -2.48 	imes 10^2 	ext{ kJ mol}^{-1}$.

The negative sign here is crucial—it tells us that the process is **exothermic**, meaning heat is being released by the system into the surroundings. The value itself means that when exactly one mole of graphite burns completely, it releases $248 	ext{ kJ}$ of heat.

### The Molar Bridge

We need to find out how much heat is generated when just $1 	ext{ g}$ of this powder is burned. To connect moles to grams, we need the molar mass.

Graphite is simply an allotrope of carbon. And we know the molar mass of carbon-12 is exactly $12 	ext{ g mol}^{-1}$. This is our bridge to solve the problem. It tells us that $12 	ext{ g}$ of graphite is equivalent to $1 	ext{ mole}$.

### The Unitary Method

Since one mole, which is $12 	ext{ g}$, releases $248 	ext{ kJ}$ of heat, we can use the simple unitary method to find the heat released by a single gram.

The heat generated by one gram will simply be the total heat per mole divided by the molar mass:

$$	ext{Heat per gram} = \frac{248 	ext{ kJ}}{12 	ext{ g}}$$

Let's do the math. Dividing $248$ by $12$ gives us $20.66 	ext{ kJ}$. This is the exact amount of heat released by burning one gram of graphite.

### The Final Polish

The question specifically asks for the answer to the **nearest integer**.

Looking at our result, $20.66$ is closer to $21$ than to $20$ because the decimal part $.66$ is greater than $.5$. So, we round it off to $21$.

And that is our final answer! A simple application of stoichiometry and thermochemistry yields a clean, integer result.

Animated Solution for Chemistry - Chemical Thermodynamics: If the standard molar enthalpy change for combustion of graphite powder is $- 2.48 \times 1 0^{2} kJ mol^{- 1}$ , the amount of heat generated on combustion of $1 g$ of graphite powder is ...... kJ. (Nearest integer)

Enter Numerical Value:

Visualized Solution

\text{Understanding the Given Data}

\text{Molar Mass of Graphite}

\text{Heat Generated per Gram}

\text{Calculation}

\text{Rounding Off}

\text{Food for Thought}

The Sigma Insight: Enthalpy and Hess's Law

The Burning Question

The Molar Bridge

The Unitary Method

The Final Polish

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