Question

How much heat is produced by burning 4.00 moles of acetylene under standard state conditions?

Answer 1

Combustion of acetylene is represented as:

C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l)

ΔHrxn = ∑nΔH°f(products) - ∑nΔH°f(reactants)

n = # moles

ΔH°f = standard enthalpy of formation from thermochemical tables

ΔHrxn = [2ΔH°f(CO2)+ΔH°f(H2O)] - [ΔH°f(C2H2)+ 3/2ΔH°f(O2)]

          = [2(-393.5)-285.8] - [226.7 + 0] = -1299.5 kJ/mol

Based on the above:

1 mole of acetylene produces -1299.5 kJ of heat

Therefore, heat produced by burning 4 moles is= -1299.5 *4 = -5198 kJ

Ans: -5198 kJ of heat is produced by burning 4.00 moles of acetylene under standard state conditions

   

Answer 2

The amount of heat produced by burning 4.00 moles of acetylene under standard state conditions is 6786.14 kJ.

Given the following data;

• Number of moles of acetylene (STP) = 4.00 moles.

To find the amount of heat produced by burning 4.00 moles of acetylene under standard state conditions:

First of all, we would write the properly balanced equation for the combustion (burning) of acetylene:

                   $2C_2H_2_{(g)} + 5O_2{(g)} ---> 4CO_2_{(g)} + 2H_2O_{(l)}$

Mathematically, the heat of combustion produced by burning acetylene under standard state conditions is calculated as follows:

$\Delta H_c = \Delta H_f(products) - \Delta H_f(reactants)$

Where:

• $\Delta H_c$ is the heat of combustion.

• $\Delta H_f$ is the heat of formation.

$\Delta H_c = [4\Delta H_f(C0_2) + 2\Delta H_f(H_20)] - [2\Delta H_f(C_2H_2) + 5\Delta H_f(0_2)]$

Substituting the values, we have:

$\Delta H_c = [4(-393.509) + 2(-285.8)] - [2(226.8) + 5(0)]\\\\\Delta H_c = [1574.036 + 576.1] - [453.6 + 0]\\\\\Delta H_c = 2150.136 - 453.6\\\\\Delta H_c = 1696.536 \;kJ/mol$

1 mole of acetylene = 1696.536 kJ/mol

4 moles of acetylene = X kJ/mol

Cross-multiplying, we have:

$X = 4 \times 1696.536$

X = 6786.14 kJ

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