Question

The following two reactions are known ${Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g); \Delta H = - 26.8 \, kJ }$ ${FeO(s) + CO (g) -> Fe(s) + CO2(g); \Delta H - 16.5 \, kJ}$ The value of $\Delta H$ for the following reaction ${Fe2O3(s) + CO(g) -> 2FeO(s) + CO2(g) }$ is

• A. - 43.3 kJ
• B. -10.3 kJ
• C. +6.2 kJ
• D. +10.3 kJ

Answer 1

The Correct Option is C

Given,
(I)$F{{e}_{2}}O(s)+3CO(g)\xrightarrow[{}]{{}}2Fe(s)+3C{{O}_{2}}(g);$
$\Delta H=-26.8\,\text{kJ}$
(II)$FeO(s)+CO(g)\xrightarrow[{}]{{}}Fe(s)+C{{O}_{2}}(g);$
$\Delta H=-16.5\,\text{kJ}$
On multiplying Eq (II) with 2, we get
(III) $2FeO(s)+2CO(g)\xrightarrow[{}]{{}}2Fe(s)+2C{{O}_{2}}(g);$
$\Delta H=-33\,\text{kJ}$
On subtracting Eq (III) from I, we get
$F{{e}_{2}}{{O}_{3}}(s)+CO(g)\xrightarrow[{}]{{}}2FeO(s)+C{{O}_{2}}(g);$
$\Delta H=-26.8-(-33)$
$=+6.2\,\text{kJ}$

Answer 2

Ans: To find the value of ΔH for the given reaction:

Fe2O3(s) + CO(g) → 2FeO(s) + CO2(g)

Hess's Law, which states that the enthalpy change for a reaction is the same regardless of whether it occurs in one step or through a series of intermediate steps. 

Step1 : 

Reverse the 2nd equation: 

Fe(s) + CO2(g) → FeO(s) + CO(g) (Note that we have reversed the second reaction and changed the sign of ΔH)

Step 2: Multiply the enthalpy of the 2nd equation: 

2Fe(s) + 2CO2(g) → 2FeO(s) + 2CO(g), ΔH = 16.5x2= 33 kJ

Step 3: Now add both the equation 

ΔH(target reaction) = ΔH(step 1) + ΔH(step 2)

                                   = -26.8 + 33 

                                   = 6.2kJ