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Question

Calculate standard entropy change in the reaction
$$Fe_2O_3 (s) + 3H_2 (g) \rightarrow 2Fe (s) + 3H_2O({l})$$
Given : $$S^0_m (Fe_2 O_3, S) = 87.4, S^0_m (Fe, S) = 27.3, S^0_m (H_2, g) = 130.7, S^0_m (H_2 O, l) = 69.9 JK^{-1} mol^{-1}$$


Solution

$$\quad { Fe }_{ 2 }{ O }_{ 3 }(s)+3{ H }_{ 2 }\quad \longrightarrow 2{ Fe }_{ (s) }+3{ H }_{ 2 }O(l)$$
$$\quad \Delta S^{ o  }_{ reaction }=\Sigma \Delta { S }^{ o  }(products)-\Sigma \Delta { S }^{ o  }(reactants)$$
=$$2{ S }_{ m }^{ o }(Fe,S)+3{ S }_{ m }^{ o }({ H }_{ 2 }O,l)-[{ S }_{ m }^{ o }({ Fe }_{ 2 }{ O }_{ 3 },S)+3{ S }_{ m }^{ o }({ H }_{ 2 },g)]$$
=$$(2 \times 27.3)+(3 \times 69.9)-87.4-(3 \times 130.7)$$
$$\quad \Delta S^{ o  }_{reaction}$$=-$$215.2Jk^{ -1 }$$

Chemistry

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