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How would you calculate the standard entropy change for the combustion of methanol at 25 C? #2CH_3OH(lamda) + 3O_2(g) -> 2CO_2(g) + 4H_2O(g)#

Answer 1

Olivia Anton

You would calculate it like this:

#sf(DeltaS^@=Sigma[S^@"products"]-Sigma[S^@"reactants"])#

#sf(DeltaS^@=(2xx213.7+4xx188.8)-(2xx127.2+3xx205.1))#

#sf(DeltaS^@=1,182.6-869.7)#

#sf(DeltaS^@=+312.9color(white)(x)"J""/"K."mol")#

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Answer 2

Avery Adams

To calculate the standard entropy change ((\Delta S^\circ)), use the formula:

[ \Delta S^\circ = \sum nS^\circ_{\text{products}} - \sum mS^\circ_{\text{reactants}} ]

Where (n) and (m) are the coefficients in the balanced chemical equation, and (S^\circ) is the standard molar entropy. Plug in the values for the standard molar entropy for each substance at 25°C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How would you calculate the standard entropy change for the combustion of methanol at 25 C? #2CH_3OH(lamda) + 3O_2(g) -&gt; 2CO_2(g) + 4H_2O(g)#

How would you calculate the standard entropy change for the combustion of methanol at 25 C? #2CH_3OH(lamda) + 3O_2(g) -> 2CO_2(g) + 4H_2O(g)#