QuestionJuly 31, 2025

Question 21 Calculate Delta _(r)G^circ at 25.0^circ C for the reaction below. 2Na(s)+2H_(2)O(l)arrow 2NaOH(aq)+H_(2)(g) Given: Delta _(r)H^circ =-314kJ and Delta _(r)S^circ =-117J/K (Report your answer in units of kJ, to one decimal place, i.e xxx.x, and no units are necessary in the answer.) (Remember to include a negative sign, if necessary.)

Question 21 Calculate Delta _(r)G^circ at 25.0^circ C for the reaction below. 2Na(s)+2H_(2)O(l)arrow 2NaOH(aq)+H_(2)(g) Given: Delta _(r)H^circ =-314kJ and Delta _(r)S^circ =-117J/K (Report your answer in units of kJ, to one decimal place, i.e xxx.x, and no units are necessary in the answer.) (Remember to include a negative sign, if necessary.)
Question 21 Calculate Delta _(r)G^circ at 25.0^circ C for the reaction below. 2Na(s)+2H_(2)O(l)arrow 2NaOH(aq)+H_(2)(g) Given: Delta _(r)H^circ =-314kJ and Delta _(r)S^circ =-117J/K (Report your answer in units of kJ, to one decimal place, i.e xxx.x, and no units are necessary in the answer.) (Remember to include a negative sign, if necessary.)

Solution
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Answer

-279.1 Explanation 1. Convert Temperature to Kelvin T = 25.0 + 273.15 = 298.15 \, K 2. Use Gibbs Free Energy Formula **\Delta_r G^{\circ} = \Delta_r H^{\circ} - T \Delta_r S^{\circ}** 3. Calculate \Delta_r G^{\circ} Substitute the values: \Delta_r G^{\circ} = -314 \, kJ - (298.15 \, K)(-117 \, J/K \times \frac{1 \, kJ}{1000 \, J}) Simplify: \Delta_r G^{\circ} = -314 \, kJ + 34.9 \, kJ Result: \Delta_r G^{\circ} = -279.1 \, kJ

Explanation

1. Convert Temperature to Kelvin<br /> $T = 25.0 + 273.15 = 298.15 \, K$<br />2. Use Gibbs Free Energy Formula<br /> **$\Delta_r G^{\circ} = \Delta_r H^{\circ} - T \Delta_r S^{\circ}$**<br />3. Calculate $\Delta_r G^{\circ}$<br /> Substitute the values: $\Delta_r G^{\circ} = -314 \, kJ - (298.15 \, K)(-117 \, J/K \times \frac{1 \, kJ}{1000 \, J})$<br /> Simplify: $\Delta_r G^{\circ} = -314 \, kJ + 34.9 \, kJ$<br /> Result: $\Delta_r G^{\circ} = -279.1 \, kJ$
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