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Exercises · 5.11

Q.Enthalpy of combustion of carbon to CO2CO_2 is –393.5 kJ mol−1^{-1}. Calculate the heat released upon formation of 35.2 g of CO2CO_2 from carbon and dioxygen gas.

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The enthalpy of combustion tells us the heat released when one mole of carbon burns to form CO2CO_2. For 35.2 g of CO2CO_2 (0.8 mol), we scale the given enthalpy proportionally to find –314.8 kJ of heat released.

Why this approach works

Enthalpy of combustion is defined for the complete burning of one mole of a substance. When carbon burns in oxygen, the reaction is:

C(s)+O2(g)→CO2(g)ΔH=−393.5 kJ mol−1C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -393.5 \text{ kJ mol}^{-1}

The negative sign tells us heat is released (exothermic). This value is per mole of carbon consumed, which is the same as per mole of CO2CO_2 formed because the stoichiometry is 1:1.

The question asks for heat released when a specific mass of CO2CO_2 forms. The strategy is simple: find how many moles of CO2CO_2 we're making, then scale the enthalpy change proportionally.

Step-by-step solution

1. Find the molar mass of CO2CO_2

Carbon has atomic mass 12 g mol−1^{-1}, oxygen 16 g mol−1^{-1}.

MCO2=12+2(16)=44 g mol−1M_{CO_2} = 12 + 2(16) = 44 \text{ g mol}^{-1}

2. Calculate moles of CO2CO_2 formed

Given mass is 35.2 g.

n=massM=35.244=0.8 moln = \frac{\text{mass}}{M} = \frac{35.2}{44} = 0.8 \text{ mol}

3. Scale the enthalpy change

The combustion of 1 mol of carbon releases 393.5 kJ. For 0.8 mol:

ΔH=0.8×(−393.5)=−314.8 kJ\Delta H = 0.8 \times (-393.5) = -314.8 \text{ kJ}

The negative sign confirms heat is released to the surroundings. …

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