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What is the standard enthalpy of reaction (in kJ) when two moles of Fe ₂ O ₃(s) reacts with H ₂ gas to give Fe metal? H_f^ of Fe ₂ O ₃(s) and H ₂ O (l) are -824.2 and -285.83 ~kJ ~mol ⁻¹ respectively.

Options

  1. A-66.58
  2. B-33.3
  3. C-538.37
  4. D-1110.03

Correct answer

A. -66.58

Step-by-step solution

When Fe ₂ O ₃(s) reacts with H ₂(g) , it will give Fe and H ₂ O as follows: Fe ₂ O ₃+3 H ₂ 2 Fe +3 H ₂ O When 2 moles of Fe ₂ O ₃ are used, the reaction becomes as follows: 2 [ Fe ₂ O ₃+3 H ₂ 2 Fe +3 H ₂ O ] Given, H_f^ for Fe ₂ O ₃(s)=-824.2 ~kJ ~mol ⁻¹ and H_f^ for H ₂ O (l)=-285.83 ~kJ ~mol ⁻¹ Standard enthalpy of reaction ( H_R^ )= H_f^ (Product) - H_f^ (Reacant ) and . array c H_f^ for H ₂=0 H_f^ for Fe =0 array ] (as are present in free state) Thus, aligned H_R^ & =[6 -285.83]-[2 (-824.2)] H_R^ & =(-1714.98)+

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