Question

Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.04 moles of CO(g) react at standard conditions. S°system = J/K g

Answer 1

6.29 J/(g*K)

Step-by-step explanation:

Hi, you haven't provided the equation that we need for the calculation of entropy change. However, the equation below will be used for the calculation:

`2CO_((g)) + O_(2(g)) ---2CO_(2(g))`

Δ
`s_(rxn) =` 2*Δ
`s_(f)(CO_(2(g)))`] - [2*Δ
`s_(f)(CO_((g)))` + Δ
`s_(f)(O_(2(g)))`]

Δ
`s_(rxn)` is the standard entropy change for the reaction [J/(mol.K)]

Δ
`s_(f)(CO_(2(g)))` is the standard entropy of formation of carbon dioxide gas = 213.79 J/(mol.K)

Δ
`s_(f)(CO_((g)))` is the standard entropy of formation of carbon monoxide gas = 197.66 J/(mol.K)

Δ
`s_(f)(O_(2(g)))` is the standard entropy of formation of oxygen gas = 205.03 J/(mol.K)

Therefore,

Δ
`s_(rxn)` = [2*213.79] - [2*197.66+205.03] = - 172.77 J/(mol*K)

From the given chemical reaction, when 2 moles of carbon monoxide gas reacts, the entropy change for the reaction -172.77 J/(mol*K). Therefore, when 2.04 moles of carbon monoxide gas reacts, the entropy change for the reaction is -172.77*2.04/2 = -176.23 J/(mol*K).

Thus, converting J/(mol*K) to J/(g*K):

molar mass of carbon monoxide is 28 g/mol.

Thus, the entropy change in J/(g*K) = -176.23/28 = 6.29 J/(g*K)