The reaction of cyanamide, NH2CN(s) with oxygen was run in a bomb calorimeter and ΔU was found to be

Q: Why is the heat measured in a bomb calorimeter equal to $\Delta U$ and not $\Delta H$?

A bomb calorimeter is a sealed, rigid container, meaning the volume is strictly constant ($\Delta V = 0$). Since work done $W = -P\Delta V = 0$, the first law of thermodynamics ($\Delta U = q + W$) simplifies to $\Delta U = q_v$. Enthalpy ($\Delta H$), on the other hand, represents heat at constant pressure ($q_p$).

Q: Why do we only count gaseous moles for $\Delta n_g$?

The volume occupied by solids and liquids is negligible compared to gases. The work done by expansion or compression ($P\Delta V$) is almost entirely due to the change in the number of moles of gas. Hence, we only consider $\Delta n_g$ in the ideal gas approximation $P\Delta V = \Delta n_g RT$.

Q: I got a completely different answer. What did I do wrong?

The most common trap here is a unit mismatch! $\Delta U$ is given in $\text{kJ mol}^{-1}$, but the standard value of $R$ is $8.314 \text{ J K}^{-1} \text{ mol}^{-1}$. You must divide $R$ by $1000$ to convert it to kilojoules before adding it to $\Delta U$.

Q: Why is the water produced considered a liquid and not a gas?

The standard temperature given is $298 \text{ K}$ ($25^\circ\text{C}$). At room temperature, water exists in its stable liquid state. Always check the temperature to determine the physical state of $\text{H}_2\text{O}$!

Q: Why did we take the magnitude at the end?

The question specifically asked for the 'magnitude' of $\Delta H_{298}$. Magnitude refers to the absolute value, which strips away the negative sign indicating the exothermic nature of the reaction, leaving just the numerical size of the energy change.

Question

The reaction of cyanamide, $	ext{NH}_2	ext{CN}(s)$ with oxygen was run in a bomb calorimeter and $\Delta U$ was found to be $-742.24 	ext{ kJ mol}^{-1}$. The magnitude of $\Delta H_{298}$ for the reaction
$	ext{NH}_2	ext{CN}(s) + \frac{3}{2}	ext{O}_2(g) \longrightarrow 	ext{N}_2(g) + 	ext{CO}_2(g) + 	ext{H}_2	ext{O}(l)$
is ......... kJ (Rounded off to the nearest integer).
[Assume ideal gases and $R = 8.314 	ext{ J mol}^{-1} 	ext{ K}^{-1}$]

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are conducting a high-stakes experiment in a chemistry lab. You place a sample of cyanamide inside a heavy, sealed steel container known as a **bomb calorimeter**.

Because the walls of this container are completely rigid, the volume cannot change. This means the work done by expansion is exactly zero.

Therefore, any heat released by the combustion reaction goes entirely into changing the internal energy of the system.

This is why the heat measured in a bomb calorimeter is directly equal to $\Delta U$. The problem gives us this value as $-742.24 	ext{ kJ mol}^{-1}$.

### The Master Equation

While the bomb calorimeter gives us $\Delta U$, we are asked to find the enthalpy change, $\Delta H$.

Enthalpy represents the heat of reaction if it were to occur in an open container at constant pressure.

To bridge the gap between constant volume ($\Delta U$) and constant pressure ($\Delta H$), we use the fundamental thermodynamic relation:

$$ \Delta H = \Delta U + \Delta n_g RT $$

This equation is our master key. The term $\Delta n_g RT$ accounts for the work energy that would have been spent pushing back the atmosphere if the gases were allowed to expand.

### Decoding the Gaseous Moles

The most critical part of this problem is calculating $\Delta n_g$, which is the change in the number of moles of **gaseous** substances.

We must look at the balanced chemical equation and strictly ignore any solids or liquids.

$$ 	ext{NH}_2	ext{CN}(s) + \frac{3}{2}	ext{O}_2(g) \longrightarrow 	ext{N}_2(g) + 	ext{CO}_2(g) + 	ext{H}_2	ext{O}(l) $$

On the reactant side, cyanamide is a solid, so we ignore it. We only have $\frac{3}{2}$ moles of oxygen gas. Thus, $n_{g,R} = 1.5$.

On the product side, water is a liquid at the standard temperature of $298 	ext{ K}$, so we ignore it. We have $1$ mole of nitrogen gas and $1$ mole of carbon dioxide gas. Thus, $n_{g,P} = 1 + 1 = 2$.

Subtracting the reactants from the products gives us our change:

$$ \Delta n_g = 2 - 1.5 = 0.5 $$

### The Unit Conversion Trap

Before we rush into the final calculation, we must pause and check our units. This is where many students lose easy marks!

Our internal energy $\Delta U$ is given in **kilojoules** ($	ext{kJ}$), but the universal gas constant $R$ is given in **joules** ($8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$).

We absolutely must convert $R$ into kilojoules by dividing by $1000$.

$$ R = 8.314 	imes 10^{-3} 	ext{ kJ K}^{-1} 	ext{ mol}^{-1} $$

### Final Calculation

Now, we are ready to substitute all our carefully prepared values into the master equation.

$$ \Delta H = -742.24 + (0.5) 	imes \left(8.314 	imes 10^{-3}ight) 	imes 298 $$

First, let's evaluate the work term $\Delta n_g RT$:

$$ \Delta n_g RT = 0.5 	imes 8.314 	imes 10^{-3} 	imes 298 \approx 1.238 	ext{ kJ mol}^{-1} $$

Adding this to our internal energy:

$$ \Delta H = -742.24 + 1.238 = -741.002 	ext{ kJ mol}^{-1} $$

The question specifically asks for the **magnitude** of $\Delta H_{298}$, rounded off to the nearest integer.

Taking the absolute value removes the negative sign, and rounding $-741.002$ gives us our final, elegant answer: **$741$**.

Enter Numerical Value:

Visualized Solution

$Understanding Bomb Calorimeter$

$The Master Equation$

$Calculating Δ n_{g}$

$Unit Conversion Check$

$Substitution$

$Calculation$

$Final Answer$

$The Way Forward$

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Decoding the Gaseous Moles

The Unit Conversion Trap

Final Calculation

Similar Questions

Consider the reaction, $N_{2} + 3 H_{2} ⟶ 2 NH_{3}$ carried out at constant temperature and pressure. If $Δ H$ and $Δ E$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

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$(Δ H - Δ E)$ for the formation of carbon monoxide (CO) from its elements at $298 K$ is $(R = 8.314 JK^{- 1} mol^{- 1})$

The value of enthalpy change ( $Δ H$ ) for the reaction $C_{2} H_{5} OH (l) + 3 O_{2} (g) \to 2C O_{2} (g) + 3 H_{2} O (l)$ at $2 7^{\circ} C$ is $- 1366.5 kJ mol^{- 1}$ . The value of internal energy change for the above reaction at this temperature will be

5 moles of an ideal gas at $100 K$ are allowed to undergo reversible compression till its temperature becomes $200 K$ . If $C_{V} = 28 JK^{- 1} mol^{- 1}$ , calculate $Δ U$ and $Δ (p V)$ for this process. ( $R = 8.0 JK^{- 1} mol^{- 1}$ )

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Understanding Bomb Calorimeter

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Master Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Δng​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Unit Conversion Check

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward