Question

In which of the reactions below is ΔH not an enthalpy of formation?

Group of answer choices

Both b and c are not enthalpy of formation reactions.

2C(s) + O2(g) → 2CO(g)

Mg(s) + Cl2(g) → MgCl2(s)

Ca(s) + 1/2O2(g) → CaO(s)

C(s) + O2(g) → CO2(g)

Answer 1

Ans- In all the reaction ΔHorxn is only an Enthalpy of formation

Explanation-

Enthalpy of reaction is the defference between the net enthalpy of formation of products - net enthalpy of formation of products

i.e

ΔHorxn =  sumation of enthalpy of formation of products - sumation of enthalpy of formation of reactants

So for

a) 2C(s) + O2(g) → 2CO(g)

ΔHorxn = 2 * ΔHof (CO(g)) - [ 2 * ΔHof (C(s)) + ΔHof (O2(g))]

Now we know ΔHof of naturally available substances is zero. So here ΔHof (O2(g), ΔHof (C(s)) = 0

So

ΔHorxn = 2 * ΔHof (CO(g)) - [ 2 * 0 + 0]

= 2 * ΔHof (CO(g))

That means here ΔHorxn is Enthalpy of formation of CO(g) only.

Similallry-

b) Mg(s) + Cl2(g) → MgCl2(s)

ΔHorxn = ΔHof (MgCl2(s)) - [ΔHof (Mg(s)) + ΔHof (Cl2(g))]

Now we know ΔHof of naturally available substances is zero. So here ΔHof (Cl2(g), ΔHof (Mg(s)) = 0

So

ΔHorxn = ΔHof (MgCl2(s)) - [0 + 0]

= ΔHof (MgCl2(s))

That means here ΔHorxn is Enthalpy of formation of MgCl2(s)) only.

c)Ca(s) + 1/2O2(g) → CaO(s)

ΔHorxn = ΔHof (CaO(s)) - [ΔHof (Ca(s)) + 1/2 * ΔHof (O2(g))]

Now we know ΔHof of naturally available substances is zero. So here ΔHof (O2(g), ΔHof (Ca(s)) = 0

So

ΔHorxn = ΔHof (CaO(s)) - [0 + 1/2 * 0)]

= ΔHof (CaO(s))

That means here ΔHorxn is Enthalpy of formation of CaO(s)) only.

d)C(s) + O2(g) → CO2(g)

ΔHorxn = ΔHof (CO2(g)) - [ΔHof (C(s)) + ΔHof (O2(g))]

Now we know ΔHof of naturally available substances is zero. So here ΔHof (O2(g), ΔHof (C(s)) = 0

So

ΔHorxn = ΔHof (CO2(g)) - [0 + 0)]

= ΔHof (CO2(g))

That means here ΔHorxn is Enthalpy of formation of CO2(g) only.