{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture 17sf

# Lecture 17sf - 2SO2(g O2(g 2SO3(g(kJ-296.8 0-395.7 S 248.1...

This preview shows pages 1–4. Sign up to view the full content.

2SO 2 (g) + O 2 (g) 2SO 3 (g) H o f (kJ) -296.8 0 -395.7 H = -197.8 kJ S() 248.1 205.0 256.6 S = -188.0 G o f (kJ) -300.2 0 -371.1 H = 2(-395.7) – 2(-296.8) = -197.8 kJ S = 2(256.6) – (2(248.1) + 205.0) = -188.0 G = H - T S G =-197.8 -298.15( 1000 0 . 188 - ) = -141.7 kJ Alternative calculation of G: G = Σ n p G o f product - Σ n r G o f reactant G = 2(-371.1) - 2(-300.2) = -141.8 kJ We can also calculate G using the formula G = Σ n p G o f product - Σ n r G o f reactant This is a shortcut, and saves some time. We would get the same result by calculating H and S from the table, and then calculating G using the formula G = H - T S This more general approach, however, lets us calculate G at different temperatures, and therefore determine reaction spontaneities at different temperatures. Consider the reaction from the previous example:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2SO 2 (g) + O 2 (g) 2SO 3 (g) As already calculated: H = -197.8 kJ S = -188.0 G = -141.8 kJ G = H - T S Since H and S are both negative, at a high enough temperature, -T S will be positive enough to be greater than the negative H, and the reaction spontaneity will reversed. This is shown in the following example. At what temperature will this reaction be spontaneous in the reverse direction under standard conditions? Under standard conditions, the reaction will be at equilibrium when G = 0 G = H - T S = 0 -197.8 - T( 1000 188 - ) = 0 T = 1052 K At temperatures above 1052 K, the reaction will be spontaneous to the left under standard conditions. Note that we assumed that H and S were both independent of temperature. This is approximately true if the temperature change is not too high. G, however, is never independent of temperature but is explicitly temperature dependent. G = H - T S If H and S are both negative, G will be < 0 (spontaneous reaction) only at a relatively low temperature. At high temperatures, G will be > 0 and the reaction will be non-spontaneous (or spontaneous in the reverse direction). If H and S are both positive, G will be < 0 (spontaneous reaction) only at a relatively high temperature. At low temperatures, G will be > 0 and the reaction will be non-spontaneous (or spontaneous in the reverse direction).
CaCO 3 (s) CaO(s) + CO 2 (g) H o f (kJ) -1207 -635.1 -393.5 S() 88.70 39.75 213.6 G o f (kJ) -1128 -604.0 -394.4 H = (-635.1 -393.5) - (-1207) = +178.4 kJ S = (213.6 + 39.75) - 88.70 = +164.65 H = +178.5 kJ S = +164.65 G = 178.4 -298.15( 1000 65 . 164 ) = 129.3 kJ Alternative calculation of G: G = Σ n p G o f product - Σ n r G o f reactant G = (-394.4 -604.0) - (-1128) = 129.6 kJ Thus the reaction is non-spontaneous at 298 K. The reverse reaction would be spontaneous at 298 K. Since H and S are both positive, at a high enough temperature, -T S will be negative enough to be greater than the positive H, and the reaction spontaneity will reversed. This is shown in the following example. At what temperature will this reaction be spontaneous to the right?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 14

Lecture 17sf - 2SO2(g O2(g 2SO3(g(kJ-296.8 0-395.7 S 248.1...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online