exam1
exam1
Note that the student will need to know (memorize) all of the mathematical formulas for the exam. The exam cover page will only have constants, conversion factors, and data - no formulas.
Heating a substance: \(q = m\cdot C_{\rm s}\cdot \Delta T\)
Phase changes: \(q = m \cdot \Delta H_{\rm change}\)
Know how to calculate for various heating scenarios and phase changes. We WILL provide the heat capacities and enthalpies of change that are needed.
Clausius-Clapeyron Equation: \( \ln\left({P_2\over P_1}\right) = {\Delta H_{\rm vap}\over R}\left({1\over T_1}-{1\over T_2}\right) \)
Your thinking here is that there are 5 variables in this equation. Somehow, we will give you 4 of them and you'll calculate the last one. Also remember... all normal boiling points have vapor pressures equal to 1 atm by definition.
\(\Delta H_{\rm solution} = \Delta H_{\rm lattice} + \Delta H_{\rm hydration}\)
Remember that the lattice energy, \(\Delta H_{\rm lattice}\), is always +positive the way we use it (expanding the solid into separate molecules or ions). The hydration energy, \(\Delta H_{\rm hydration}\), is always –negative in the way we use it.
Henry's Law: \( K_{\rm H} = {P_{\rm A} \over [{\rm A}]} \) where \([{\rm A}]\) is concentration of A in solution.
Remember: "Likes Dissolve Likes"
Vapor Pressure Lowering
Raoult's Law: \(P_{\rm A} = \chi_{\rm A}\cdot P_{\rm A}^\circ \) for volatile solvent A and non-volatile solute
Raoult's Law: \(P_{\rm total} = \chi_{\rm A}\cdot P_{\rm A}^\circ + \chi_{\rm B}\cdot P_{\rm B}^\circ \) for volatile solvent A and volatile solute B
Freezing Point Depression: \(\Delta T_{\rm f} = i\cdot k_{\rm f} \cdot m \)
Boiling Point Elevation: \(\Delta T_{\rm b} = i\cdot k_{\rm b} \cdot m \)
Osmotic Pressure: \(\Pi = i\cdot MRT \)
Remember: All concentration terms will need adjustment with the van't Hoff factor, \(i\), if the solute dissociates into ions (salts).
None of this will be printed on the exam. You need to KNOW this on your own.
| salt ratio | van't Hoff factor | chemical reaction | equilibrium formula | Ksp(x) | molar solubility |
|---|---|---|---|---|---|
| \(1:1\) | \(i=2\) | \({\rm MX(s) \rightleftharpoons M^+(aq) + X^–(aq)}\) | \(K_{\rm sp} = [\rm M^+][\rm X^–]\) | \(K_{\rm sp} = x^2\) | \(x = \sqrt{K_{\rm sp}}\) |
| \(1:2\) | \(i=3\) | \({\rm MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^–(aq)}\) | \(K_{\rm sp} = [\rm M^{2+}][\rm X^–]^2\) | \(K_{\rm sp} = 4x^3\) | \(x = \root 3 \of {K_{\rm sp}\over 4}\) |
| \(1:3\) | \(i=4\) | \({\rm MX_3(s) \rightleftharpoons M^{3+}(aq) + 3X^–(aq)}\) | \(K_{\rm sp} = [\rm M^{3+}][\rm X^–]^3\) | \(K_{\rm sp} = 27x^4\) | \(x = \root 4 \of {K_{\rm sp}\over 27}\) |
| \(2:3\) | \(i=5\) | \({\rm M_2X_3(s) \rightleftharpoons 2M^{3+}(aq) + 3X^{2–}(aq)}\) | \(K_{\rm sp} = [\rm M^{3+}]^2[\rm X^{2–}]^3\) | \(K_{\rm sp} = 108x^5\) | \(x = \root 5 \of {K_{\rm sp}\over 108}\) |
| forward\(\rightarrow\) | \(Q < K\) | which means more solid will dissolve if there is any or nothing happens because the solution is undersaturated. |
| equil\(\rightleftharpoons\)ibrium | \(Q = K\) | which means the solution concentrations will hold constant and there is no net change. |
| \(\leftarrow\)reverse | \(Q > K\) | means the solution is oversaturatured and a precipitation will occur until the concentrations drop to saturation level. |