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Chemistry 13 Online
OpenStudy (fifib):

At which temperature range would the following reaction be spontaneous? 2NH4NO3(s) 2N2(g) + 4H2O(g) + O2(g) ?H = -236 kJ/mol ?S = 46 J/mol K A. The reaction is never spontaneous. B. at temperatures below 5130 K C. at temperatures above 5130 K D. The reaction is spontaneous at all temperatures.

OpenStudy (fifib):

I'm not sure how ΔG=ΔH−TΔS will help

OpenStudy (photon336):

do you know what spontaneity means?

OpenStudy (photon336):

@fifib

OpenStudy (fifib):

i mean I could look it up rn but it would be cheating lol so NO

OpenStudy (photon336):

spontaneity means DOES a reaction have a tendency to go from reactants to products. It DOES NOT tell us how fast the reaction happens. a spontaneous reaction could take a few minutes, days or even a month to happen.

OpenStudy (photon336):

Questions? if not i'll go on

OpenStudy (fifib):

what do you mean by 'GO FROM'

OpenStudy (photon336):

A + B --> C+ D So reactants combine /react chemically to form products

OpenStudy (fifib):

Oh alright yes

OpenStudy (photon336):

sp yeah we need to use \[\Delta G = \Delta H - T \Delta S \]

OpenStudy (photon336):

so actually what you can do is plug in the numbers and if delta G < 0 That means your reaction is sponataneous .

OpenStudy (fifib):

it is negative. because i put in -236-46

OpenStudy (photon336):

We start with our formula \[\Delta G = \Delta H - T \Delta S \] Then here is the information that we have \[\Delta H = -236,000 J/mol\] \[\Delta S = 46 J/mol K \] then we plug everything in. \[-236,000\frac{ J }{ mol } - T*46 \frac{ J }{ mol*K } \] if you notice, we don't have any information about the temperature. try plugging in 5130 K where temperature is.

OpenStudy (fifib):

OKAY hold on

OpenStudy (fifib):

I got -471980 and I have a feeling that's wrong :(

OpenStudy (photon336):

yeah it's definitely spontaneous

OpenStudy (fifib):

im so lost.. which one is the answer though.. D?

OpenStudy (photon336):

ill show you

OpenStudy (fifib):

Okay @Photon336

OpenStudy (anonymous):

can I help

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