Use the following data to calculate the standard heat (enthalpy) of formation, δ h°f, of manganese(iv) oxide, mno2 ( s). 2mno2( s) → 2mno( s) + o2( g), δh = 264 kj mno2( s) + mn( s) → 2mno( s), δh = -240 kj(a) -504 kj (b) -372 kj (c) -24 kj (d) 24 kj

(A)

Explanation:

The enthalpy of formation of a substance is the enthalpy of the reaction where this substance is formed by its constituents species. So, for MnO2, the enthalpy of formation is the enthalpy of the reaction:

Mn(s) + O2(g) --> MnO2(s)

By the Hess' law, when a reaction follows steps, the enthalpy of the overall reaction is the sum of the enthalpy of the steps. In this sum, the intermediaries must be canceled, so, some changes may have to be done in the reactions. If the reaction is inverted, the signal of the enthalpy inverts too, and if it's multiplied by some constant, the enthalpy is multiplied too.

2MnO2 (s) --> 2MnO(s) + O2(g) ΔH = 264 kJ (must be inverted)

MnO2(s) + Mn(s) --> 2 MnO(s) ΔH = -240 kJ

O2(g) + 2MnO(s) --> 2MnO2(s) ΔH = -264 kJ

MnO2(s) + Mn(s) --> 2 MnO(s) ΔH = -240 kJ

MnO is canceled, and 2MnO2 - MnO2 = MnO2 in the products because it was where have more of it:

O2(g) + Mn(s) --> MnO2(s)

ΔH = -264 + (-240) = -504 kJ

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24