Use the following reactions and given Δh values to find standard enthalpies of reactions (in kilojoules) given below. C(s)+O2(g)→CO2(g) ΔH= -393.5 kJ 2CO(g)+O2(g)→2CO2(g) ΔH= -566.0 kJ2H2(g)+O2(g)→2H2O(g) ΔH= -483.6 kJ

The heat of the reaction or standard enthalpy of the reaction CO(g) + H₂O(g) → CO₂(g) + H₂(g) is ΔH(rxn) = -41.2 kJ

Explanation:

The reaction whose standard enthalpy is required, as obtained from the internet is

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

The formation reaction for some of the reactants and products are given as

C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ

2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ

2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -483.6 kJ

To find the standard enthalpies of the given reaction, we need the heat of formation of each of the species involved in the reaction

ΔH (CO₂(g)) = -393.5 kJ

ΔH (H₂O(g)) = -483.6 kJ ÷ 2 = -241.8 kJ (Because 2 moles of H₂O(g) are formed in the given formation reaction)

ΔH (O₂(g)) = ΔH (H₂(g)) = 0 kJ (No heat of formation for elements)

ΔH (CO(g)) = ? (This isn't given)

But it can be calculated from the second given reaction

2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ

Heat of reaction = ΔH(products) - ΔH(reactants)

Heat of reaction = -566.0 kJ

ΔH (products) = 2 × ΔH (CO₂(g)) = 2 × -393.5 = -787 kJ

ΔH (reactants) = [2 × ΔH (CO(g))] + [1 × ΔH (O₂(g))] = 2 × ΔH (CO₂(g))

Hence, we have

-566 = -787 - [2 × ΔH (CO₂(g))]

2 × ΔH (CO₂(g)) = -787 + 566 = -221 kJ

ΔH (CO₂(g)) = -221 ÷ 2 = -110.5 kJ

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

Heat of reaction = ΔH(products) - ΔH(reactants)

ΔH (products) = [ΔH (CO₂(g))] + [ΔH (H₂(g))]

= -393.5 + 0 = -393.5 kJ

ΔH (reactants) = [1 × ΔH (CO(g))] + [1 × ΔH (H₂O(g))] = -110.5 - 241.8 = -352.3 kJ

Heat of reaction = -393.5 - (-352.3) = -41.2 kJ

Hope this Helps!!!

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3