Calculate the enthalpy change for the reaction, Fe2O3+3CO arrow 2Fe+3CO2; from the following data: 2Fe+(3/2)O2 arrow Fe2O3;ÎH=-177.1 kcal C+(1/2)O2 arrow CO;ÎH=-32.8 kcal C+O2 arrow CO2;ÎH=-94.3 kcal Express the magnitude of your answer by rounding off to one significant figure.

Question

Calculate the enthalpy change for the reaction, Fe2O3+3CO arrow 2Fe+3CO2; from the following data: 2Fe+(3/2)O2 arrow Fe2O3;ÎH=-177.1 kcal C+(1/2)O2 arrow CO;ÎH=-32.8 kcal C+O2 arrow CO2;ÎH=-94.3 kcal Express the magnitude of your answer by rounding off to one significant figure. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let the enthalpy of the given reaction be

△ H_{r}

(F e)_{2} O_{3} (s) + 3 CO (g) \to 2 F e (s) + 3 (CO)_{2} (g) \dots (i)

Given equation:

2 F e (s) + \frac{3}{2} O_{2} (g) \to (F e)_{2} O_{3} (s); Δ H = - 177.1 k c a l . \dots

(ii)

C (s) + \frac{1}{2} O_{2} (g) \to CO (g); Δ H = - 32.8 k c a l . \dots (iii)

C (s) + O_{2} (g) \to (CO)_{2} (g); Δ H = - 94.3 k c a l \dots (i v)

Enthalpy is an extensive property.
Equation (i) can be obtained from equation (ii), (iii) and (iv) as given below:
Equation

(i v) \times 3 -

Equation

(iii) \times 3 -

Equation

(ii)

\equiv

Equation (i)
Thus, according to Hess' law,

[(- 98.3) \times 3] - [(- 32.8) \times 3] - [- 177.1] = Δ H_{r}

Δ H_{r} = - 282.9 + 98.4 + 177.1 = - 282.9 + 275.5

Δ H_{r} = - 7.4 k c a l

Thus, the enthalpy of equation (i) is

- 7.4 k c a l

.

Q. Calculate the enthalpy change for the reaction, Fe2​O3​+3CO→2Fe+3CO2​; from the following data: 2Fe+23​O2​→Fe2​O3​;ΔH=−177.1kcal C+21​O2​→CO;ΔH=−32.8kcal C+O2​→CO2​;ΔH=−94.3kcal Express the magnitude of your answer by rounding off to one significant figure.