Question

The standard enthalpy change for the following reaction is -774 kJ at 298 K. Fe(s) + 5 C(s.graphite) + 5/2 O2(g) Fe(CO)(1) AH--774 kJ What is the standard enthalpy change for the reaction at 298 K? 2 Fe(s) + 10 C(s.graphite) +5 O2(g) → 2 Fe(CO)s(1) Submit Answer

Answer 1

An extensive property is defined as the property that depends on the amount of matter involved.

Enthalpy is an extensive property.

The standard enthalpy of the reaction is as follows:

Fe(s) + 5C(s,graphite) + 5/2O2(g) $\rightarrow$ Fe(CO)5(l) $\Delta$ Ho = -774 kJ

If the amount of the matter is doubled, the enthalpy change value will be 2 times the initial enthalpy.

2Fe(s) + 10C(s,graphite) + 5O2(g) $\rightarrow$ 2Fe(CO)5(l) $\Delta$ Ho = -774 kJ x 2

Thus,

$\Delta$Ho = -774 kJ x 2

$\Delta$Ho = -1548 kJ Or -1550 kJ [3S.F]