Question

Sulfur reacts with oxygen to form sulfur dioxide (SO2(g), Delta.Hf = –296.8 kJ/mol) according to the equation below. Upper S (s) plus upper O subscript 2 (g) right arrow upper S upper o subscript 2 (g). What is the enthalpy change for the reaction? Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants.

Answer 1

-296.8

Step-by-step explanation:

Answer 2

The enthalpy change for the reaction is -296.8 kJ.

Step-by-step explanation:

S(s) + O₂(g) ⇄ SO₂ (g)

We have to find the enthalpy change by finding the difference between the sum of enthalpies of products and that of reactants along with their coefficient terms. It is termed as ΔH.

It is found that,

`\Delta H_(f^(\circ))\left(S O_(2(g))\right)=-296.8 \mathrm{kJ} / \mathrm{mol}`

`\Delta H_(f^(\circ))\left(S_((s))\right)=0 k J / m o l`

`\Delta H_(f^(\circ))\left(O_(2) _((g))\right)=0 k J / m o l`

The enthalpy change can be found out as,

`\Delta H_(r x n)=\left(1 * \Delta H_(f)\left(S O_(2)\right)_((g))\right)-\left(1 * \Delta H_(f)(S)_((s))+1 * \Delta H_(f)\left(O_(2)\right)(g)\right)`

= (1 × -296.8) - ((1 × 0) + ( 1 × 0 ))

= -296.8 kJ

So the enthalpy change for the reaction is -296.8 kJ/ mol.