Question

The reaction between phosphorous,P4(s), and chlorine, Cl2(g), is exothermic and leads to either PCl3(g) or PCl5(g) depending on the stoichiometric amount of Cl2(g) used. Given the following two chemical equations (EQ 1 and EQ 2) and their associated enthalpy changes for the formation of PCl3(g) and PCl5(g);
EQ 1: P4(s) + 6Cl2(g) → 4PCl3(g) = -2439 kJ/mol-rxn
EQ 2: P4(s) + 10Cl2(g) → 4PCl5(g) = -3438 kJ/mol-rxn
EQ 3: PCl5(g) → PCl3(g) + Cl2(g) = ?
a. Calculate the expected enthalpy change for the decomposition of one mole of PCl5(g) shown in equation 3 (EQ 3).
b. Calculate the expected enthalpy change for the decomposition of 10.50 g of PCl5(g) shown in equation 3 (EQ 3).

Answer 1

Answers on Question #55559 – Chemistry - Other

Question:

The reaction between phosphorous, $P_{4(s)}$, and chlorine, $Cl_{2(g)}$, is exothermic and leads to either $PCl_{3(g)}$ or $PCl_{5(g)}$ depending on the stoichiometric amount of $Cl_{2(g)}$ used. Given the following two chemical equations (EQ 1 and EQ 2) and their associated enthalpy changes for the formation of $PCl_{3(g)}$ and $PCl_{5(g)}$:

EQ 1: $P_{4(s)} + 6Cl_{2(g)} \rightarrow 4PCl_{3(g)} = -2439\ \mathrm{kJ/mol}$

EQ 2: $P_{4(s)} + 10Cl_{2(g)} \rightarrow 4PCl_{5(g)} = -3438\ \mathrm{kJ/mol}$

EQ 3: $PCl_{5(g)} \rightarrow PCl_{3(g)} + Cl_{2(g)} = ?$

a. Calculate the expected enthalpy change for the decomposition of one mole of $PCl_{5(g)}$ shown in equation 3 (EQ 3).

b. Calculate the expected enthalpy change for the decomposition of $10.50\ \mathrm{g}$ of $PCl_{5(g)}$ shown in equation 3 (EQ 3).

Answer:

a) $4PCl_{5(g)} \rightarrow P_{4(s)} + 10Cl_{2}$ $\Delta H = 3438\ \mathrm{kJ/mol}$

+

$P_{4(s)} + 6Cl_{2(g)} \rightarrow 4\ PCl_{3(g)}$ $\Delta H = -2439\ \mathrm{kJ/mol}$

=

$4PCl_{5(g)} \rightarrow 4PCl_{3(g)} + 4\ Cl_{2(g)}$ $\Delta H = 999\ \mathrm{kJ/mol}$

$PCl_{5(g)} \rightarrow PCl_{3(g)} + Cl_{2(g)}$ $\Delta H = 249.75\ \mathrm{kJ/mol}$

b) $Q = -\Delta H \cdot v$

$M(PCl_{5}) = 208.24\ \mathrm{g/mol}$

$v(PCl_{5}) = \frac{m}{M}$

$v(PCl_{5}) = \frac{10.50}{208.24} = 0.05\ \mathrm{mol}$

$Q(PCl_{5(g)}) = -249.75 \cdot 0.05 = -12.59\ \mathrm{kJ}$

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