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Given: i. `2Fe(s) +(3)/(2)O_(2)(g) rarr Fe_(2)O_(3)(s), DeltaH^(Theta) =- 193.4 kJ` ii. `Mg(s) +(1)/(2)O_(2)(g) rarr MgO(s), DeltaH^(Theta) =- 140.2kJ` What is `DeltaH^(Theta)`of the reaction? `3Mg +Fe_(2)O_(3)rarr 3MgO +2Fe`A. `-227.2kJ`B. `-272.3kJ`C. `227.2kJ`D. `272.3kJ` |
Answer» Substracting equation (i) from equation and (ii) multiplied by `3`, we get i. `2Fe(s) +(3)/(2) O_(2)(g) rarr Fe_(2)O_(3)(g), DeltaH^(Theta) =- 193.4kJ` ii. `3Mg(s) +(3)/(2) O_(2)(g)rarr 3MgO(s), DeltaH =- 420.6 kJ` Subtracting (i) from (ii), `3Mg(s) + Fe_(2)O_(3) rarr 3MgO + 2Fe`, `DeltaH^(Theta) =- 420.6 -(-193.4) =- 227.2kJ` |
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