What two terms are used to describe the temperature at which solids and liquids of the same substance exist at the same time?

1 answer:

4 0

<span>The answer is melting and freezing point. At the melting point the solid and liquid stage occur in equilibrium. The melting point of a material hinge on pressure and is typically quantified at standard pressure. When measured as the temperature of the opposite alteration from liquid to solid, it is denoted as the freezing point.</span>

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C2H4(g) + H2(g) → C2H6(g) ΔH = –137.5 kJ; ΔS = –120.5 J/K Calculate ΔG at 25 °C and determine whether the reaction is spontaneou

Katen [24]

Answer:-ΔG=-101.5KJ

Explanation:We have to calculate ΔG for the reaction so using the formula given in the equation we can calculate the \Delta G for the reaction.

We need to convert the unit ofΔS in terms of KJ/Kelvin as its value is given in terms of J/Kelvin

Also we need to convert the temperature in Kelvin as it is given in degree celsius.

$\Delta H=-137.5\\ \Delta S=-120J/K\\ \Delta S=-0.120KJ/K\\ T=25^{.C}\\ T=273+25=298 K\\ \Delta G=?\\ \Delta G=\Delta H-T\Delta S\\\Delta G=-137.5KJ-(278\times -0.120)\\ \Delta G=-137.5+35.76\\\Delta G=-101.74\\\Delta G=Negative$

After calculating forΔG we found that the value ofΔG is negative and its value is -101.74KJ

For a reaction to be spontaneous the value of \Delta G \ must be negative .

As the ΔG for the given reaction is is negative so the reaction will be spontaneous in nature.

In this reaction since the entropy of reaction is positive and hence when we increase the temperature term then the overall term TΔS would become more positive and hence the value of ΔG would be less negative .

Hence the value of ΔG would become more positive with the increase in temperature.

So we found the value of ΔG to be -101.74KJ

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