The standard enthalpy change for the reaction of so3(g) with h2o(l) to yield h2so4(aq) is δh∘ = -227.8 kj . use the following information s(s)+o2(g)→so2(g), δh∘ = -296.8kj so2(g)+12o2(g)→so3(g) , δh∘ = -98.9kj to calculate δh∘f for h2so4(aq) (in kilojoules per mole). [for h2o(l),δh∘f = -285.8kj/mol].

-909.3KJ/mole

Explanation:

The heat of reaction is accessible from the heat of formation of reactants and products using the formula below:

ΔH = Σ ΔHf products - Σ ΔHf reactants

Before we proceed, it is important to know that the enthalpy of formation of element is zero ,be it a single element or a molecule of an element.

From the reaction for the formation of sulphuric acid, we know we need to know the heat of formation of sulphur (vi) oxide and water. The examiner is quite generous and have us for water already.

Now we need to calculate for sulphur (vi) oxide. This is calculated as follows:

We first calculate for sulphur(iv)oxide. This can be obtained from the reaction between sulphur and oxygen. The calculation goes thus:

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole suphur(iv) oxide × x] - [ (1 mole of elemental sulphur × 0) + (1 mole of elemental oxygen × 0]

We were already told this is equal to -296.8KJ. Hence the heat of formation of sulphur(iv) oxide is -296.8KJ.

We then proceed to the second stage.

Now, here we have 1 mole sulphur (iv) oxide reacting with 0.5 mole oxygen molecule.

We go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphur (vi) oxide × y] - [ (1 mole of sulphur (iv) oxide × -296.8) + (0.5 mole of oxygen × 0)].

We already know that the ΔH here equals -98.9KJ.

Hence, -98.9 = y + 296.8

y = -296.8KJ - 98.9KJ = -395.7KJ

We now proceed to the final part of the calculation which ironically comes first in the series of sentences.

Now, we want to calculate the standard heat of formation for sulphuric acid. From the reaction, we can see that one mole of sulphur (vi) oxide, reacted with one mole of water to yield one mole of sulphuric acid.

Mathematically, we go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphuric acid × z] - [( 1 mole of sulphur vi oxide × -395.7) + ( 1 mole of water × -285.8)].

Now, we know that the ΔH for this particular reaction is -227.8KJ

We then proceed to to open the bracket.

-227.8 = z - (-395.7 - 285.8)

-227.8 = z - ( -681.5)

-227.8 = z + 681.5

z = -227.8-681.5 = -909.3KJ

Hence, ΔH∘f for sulphuric acid is -909.3KJ/mol

c.

explanation:

plasma is the state of matter where the substance is ionized and has no definite shape or volume. the fact of being ionized makes the substance a good electrical conductor.