Question

Consider the following reaction at constant P. Use the information here to determine the value of ΔSsurr at 355 K. Predict whether or not this reaction will be spontaneous at this temperature. 2NO(g) + O2(g) → 2NO2(g) ΔH = -114 kJ Consider the following reaction at constant . Use the information here to determine the value of at 355 . Predict whether or not this reaction will be spontaneous at this temperature. 2(g) + (g) 2(g) = -114 ΔSsurr = +114 kJ/K, reaction is spontaneous ΔSsurr = -114 kJ/K, reaction is spontaneous ΔSsurr = -321 J/K, reaction is spontaneous ΔSsurr = +321 J/K, reaction is spontaneous ΔSsurr = +114 J/K, it is not possible to predict the spontaneity of this reaction without more information.g

Answer 1

The given reaction is as follows:

2NO (g) + O₂ (g) = 2NO₂ (g), ΔH = -114 kJ

It is known that dSsurr = -dHsys / T (Temp = 355 K)

So,  dSsurr = - (-114 × 1000) / 355

dSsurr = +321.12 J/K

Hence, the value of dSsurr is +321.12 J/K

For a reaction to be spontaneous, dG<0,

Also dStotal = dSsys + dSsurr > 0

It is known that dG = dHsys - TdSsys,

Now let us assume,

dG<0

Also, dStotal = dSsys + dSsurr > 0

(-114 × 1000) - (355 × dSsys) <0

355 × dSsys > -114 × 1000

dSsys > -321

dSsys >dSsurr

dSsys + dSsurr > 0

dStotal > 0

Thus, the assumption is correct, and the given reaction is spontaneous. Hence, the final answer is Ssurr = +321 J/K reaction is spontaneous.

Answer 2

A spontaneous reaction favours the product formation of the reaction. The given reaction is spontaneous and the surrounding entropy is +321 J/K.

What is the entropy change?

Entropy change of the surrounding is the amount of energy that gets released in the surrounding environment as heat in the ratio to the temperature of the outer environment.

The balanced chemical reaction can be shown as:

$\rm 2NO (g) + O_{2} (g) \rightarrow 2NO_{2} (g), \Delta H = -114 kJ$

Change in the entropy of the surrounding can be given as:

$\rm \Delta S_{surr} = \dfrac{-\Delta H_{sys}}{T}$

Here, the change in enthalpy is -114 kJ and the temperature is 355 K.

Substituting values in the equation above:

$\begin{aligned} \rm \Delta S_{surr} &= \dfrac{- (-114 \times 1000)}{355}\\\\&= -321.12 \;\rm J/K\end{aligned}$

Thus, the entropy of the surrounding is +321.12 J/K.

Now to check whether the reaction is spontaneous or not it must be $\rm \Delta G < 0.$

The total entropy is given as:

$\rm \Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$

And,

$\rm \Delta G = \Delta H_{sys} - T\Delta S_{sys}$

If $\rm \Delta G < 0$ then,

$\begin{aligned}\rm \Delta S_{total} &= \rm \Delta S_{sys} + \Delta S_{surr} > 0\\\\& = (-114 \times 1000) - (355 \times \rm \Delta S_{sys}) < 0\\\\& = \rm \Delta S_{sys} > -321\end{aligned}$

Solving further:

$\begin{aligned}&= \rm \Delta S_{sys} > \Delta S_{surr} \\\\& = \rm \Delta S_{sys} + \Delta S_{surr} > 0\\\\& = \rm \Delta S_{total} > 0\end{aligned}$

Therefore, the reaction is spontaneous and the surrounding entropy is +321 J/K.

Learn more about entropy here:

https://brainly.com/question/1301642