(ii) Predict the entropy changes for each of the equations given in part (i) Q 3(b) [10 Marks] When 88 grams CO2 (g) is heated at a constant pressure of 1.25 atm. Its temperature increases from 250 K to 277K. The molar heat capacity at constant pressure, Cp m For CO2 is 37.11 J K-1 mol-1. Using this information, calculate (i) The heat transferred, q (ii) The changes in Enthalpy, ?H (iii) The change in internal energy, ?U Q 3(c) [2.5 Marks] Calculate the entropy change (?S, J K-1 mol-1) when 2.00 mol of mercury vapour at its boiling point (Tb = 356.55 °C) condenses. (in the process the temperature does not change, for mercury ?Hvap is + 59.3 kJ mol-1).
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(i) The heat transferred, q, can be calculated using the formula q = nCpΔT, where n is the number of moles, Cp is the molar heat capacity at constant pressure, and ΔT is the change in temperature. Show more…
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(a) Calculate the change in entropy when 1.00 mol of water (molecular mass 18.0 $\mathrm{g} / \mathrm{mol} )$ at $100^{\circ} \mathrm{C}$ evaporates to form water vapor at $100^{\circ} \mathrm{C}$ . (b) Repeat the calculation of part (a) for 1.00 $\mathrm{mol}$ of liquid nitrogen, 1.00 $\mathrm{mol}$ of silver, and 1.00 $\mathrm{mol}$ of mercury when each is vaporized at its normal boiling point. (See Table 17.4 for the heats of vaporization, and Appendix D for the molar masses. Note that the nitrogen molecule is $\mathrm{N}_{2 \cdot}$ ) (c) Your results in parts (a) and (b) should be in relatively close agreement. (This is called the rule of Drepez and Trouton.) Explain why this should be so, using the idea that entropy is a measure of the randomness of a system.
Given the following values for the changes in enthalpy $(\Delta H)$ and entropy $(\Delta S),$ which of the following processes can take place at $298 \mathrm{K}$ without violating the Second Law of Thermodynamics? (a) $\Delta H=-84 \mathrm{kJ} \mathrm{mol}^{-1}\left(-20 \mathrm{kcal} \mathrm{mol}^{-1}\right)$ $\Delta S=+125 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\left(+30 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$ (b) $\Delta H=-84 \mathrm{kJ} \mathrm{mol}^{-1}\left(-20 \mathrm{kcal} \mathrm{mol}^{-1}\right)$ $\Delta S=-125 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\left(-30 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$ (c) $\Delta H=+84 \mathrm{kJ} \mathrm{mol}^{-1}\left(+20 \mathrm{kcal} \mathrm{mol}^{-1}\right)$ $\Delta S=+125 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\left(+30 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$ (d) $\Delta H=+84 \mathrm{kJ} \mathrm{mol}^{-1}\left(+20 \mathrm{kcal} \mathrm{mol}^{-1}\right)$ $\Delta S=-125 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\left(-30 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$
A Standard entropy changes cannot be measured directly in the laboratory. They are calculated from experimentally obtained values of $\Delta G^{0}$ and $\Delta H^{0} .$ From the data given here, calculate $\Delta S^{0}$ at $298 \mathrm{~K}$ for each of the following reactions. $$ \text { (a) } \mathrm{OF}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) $$ $$ \begin{array}{lr} \Delta H^{0}=-323.2 \mathrm{~kJ} / \mathrm{mol} & \Delta G^{0}=-358.4 \mathrm{~kJ} / \mathrm{mol} \\ \text { (b) } \mathrm{CaC}_{2}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s})+\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \\ \Delta H^{0}=-125.4 \mathrm{~kJ} / \mathrm{mol} & \Delta G^{0}=-145.4 \mathrm{~kJ} / \mathrm{mol} \\ \text { (c) } \mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \\ \Delta H^{0}=81.5 \mathrm{~kJ} / \mathrm{mol} & \Delta G^{0}=-26.20 \mathrm{~kJ} / \mathrm{mol} \end{array} $$
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