In Class Work

Unit 1 - Energy

2025-02-04

1

$$\begin{array}{rl}& q=mc\Delta T\\ & q=80.0(0.90)\times 34\\ & =2448g/J\end{array}$$

12

… mercury shc 0.14g/J

$$\begin{array}{rlr}& & \\ & m=\frac{Q}{c\Delta T}& \\ & =& \end{array}$$

2025-02-06

375g -15C → gas @ 125C

125—15
$\Delta T=140$
$mc\Delta T$

$$\begin{array}{rl}& M_{h_{2}o}=18.02\\ & m(h_{2}o)=20.8102108768mol\\ & \\ & \\ & -15\ast \end{array}$$ $$11.251+40.68015_2+156.9_3+274.692375_4+18.9375_5$$

2025-02-07

38g sample of ice at -11c is placed into 214g of water at 56c

find final temp
water: $mc\Delta T$
ice: $heatup-melt-heatup$

where $1$ is 56C water and $2$ is the warmed ice

$$\begin{array}{r}\end{array}mc\Delta T_1=mc\Delta T_2+(m\ast L)+mc\Delta T_{ice}$$ $$\begin{array}{rl}& (214)(4.184)56-(T_f)=(38)2.108(11)+333(38)+4.184(38)(T_f-0c)\\ & \\ & 50141.056-T_f=881.144+12654+158.99200\ast T_f\\ & 50141.056-T_f=13535.144+158.99200\ast T_f\end{array}$$

34.7c

2

5.022g of NaH reacts

5.022 / 84.007 = 0.0597807326

80 * 4.19 * 18.6 - 28.4

= -3284.96

$\Delta H=-\frac{q}{n}$

3284.96 / 0.0597807326
55kJ/mol

Why is the

2025-02-10

$$\begin{array}{rl}& 6Fe{O}_s+6C{O}_g\to 6F{e}_s+6C{o}_2;=-104\Delta H\\ & \Delta H-17.3333333\end{array}$$

Unit 2

$m{g}^{2+}+C{l}^{-}+C{l}^{-}\to MgC{l}_2$

$$r\alpha [M{g}^{2+}][C{l}^{-}][C{l}^{-}]$$ $$r\alpha [M{g}^{2+}][C{l}^{-}{]}^2$$

l

$[A]\ast 4r\ast 24^n=2n=1/2$

$$\begin{array}{rl}& {\left(\frac{0.1}{0.2}\right)}^n=\frac{4}{16}\\ & n=2\end{array}$$

C
1-4

$$\begin{array}{rl}& {\left(\frac{0.1}{0.3}\right)}^n=\frac{4}{12}\\ & n=1\end{array}$$ $$\begin{array}{rl}& \frac{4.0mol}{L}=k(0.1mol{)}^{1/2}\ast {\left(\frac{0.1mol}{L}\right)}^2\ast (0.1mol{)}^1\\ & =\frac{4.00mol}{L}=k0.00031622776\\ & =k=12649.1108813\\ & =k=1.3\ast 10^4\end{array}$$ $$\begin{array}{r}k=\frac{L^{2.5}}{mo{l}^{2.5}s}\end{array}$$

2025-02-28

Reaction Steps

$$\begin{array}{rl}1.& 2\text{NO}(g)\rightleftharpoons {\text{N}}_2{\text{O}}_2(g)\text{(fast)}\\ 2.& {\text{N}}_2{\text{O}}_2(g)+{\text{H}}_2(g)\to {\text{N}}_2\text{O}(g)+{\text{H}}_2\text{O}(g)\text{(slow)}\\ 3.& {\text{N}}_2\text{O}(g)+{\text{H}}_2(g)\to {\text{N}}_2(g)+{\text{H}}_2\text{O}(g)\text{(fast)}\end{array}$$

Overall Reaction

$2\text{NO}(g)+2{\text{H}}_2(g)\to {\text{N}}_2(g)+2{\text{H}}_2\text{O}(g)$

Rate Law

Since Step 2 is the rate-determining step:

$\text{Rate}=k[{\text{N}}_2{\text{O}}_2][{\text{H}}_2]$

Expressing $[{\text{N}}_2{\text{O}}_2]$ in terms of $[\text{NO}]$ from Step 1:

$\text{r}=k[\text{NO}{]}^2[{\text{H}}_2]$

2025-03-04

$$\begin{array}{r}a+b\to f+c\end{array}$$

$$3A\to B$$ $$Rate=k[A{]}^2[B]$$ $$O_3+2N{O}_2\to N_2{O}_5+O_2$$ $$Rate=k[O_3{]}^1[N{O}_2{]}^1$$

I = Catelyist

IO = Intermid

$$2H_2{O}_2\to 2H_{2}O+O_2$$

$Rate=k_1[H_2{O}_2{]}^1[I^{-}{]}^1$
$Rate=k[H_2{O}_2{]}^1$

2025-03-24

$$\begin{array}{rl}& R=\frac{[C{l}_2{]}^{\frac{1}{2}}[NO]}{[NOCl]}\end{array}$$

1. The equilibrium constant for the equilibrium CO(g) + H2O(g)⇌CO2(g) + H2(g) is 302 at 600k . What is the value of the equilibrium constant for the reverse reaction at the same temperature

3

PCL5 <> PCL3 + CL2
At the equilibrium point in the decomposition of phosphorus pentachloride to chlorine and phosphorus trichloride, the following concentrations are obtained: 0.010 mol/L PCl5, 0.15 mol/L PCl3 and 0.37 mol/L Cl2. Determine the Keq for the reaction

$$\begin{array}{rl}& K_{eq}=\frac{[PC{l}^3][C{l}^2]}{[PC{l}^5]}\\ & K_{eq}=\frac{\left[0.15\frac{mol}{L}\right]\left[0.37\frac{mol}{L}\right]}{0.010\frac{mol}{L}}\\ & K_{eq}=\frac{0.0555}{0.010}\\ & K_{eq}=5.55\end{array}$$

4

The colourless gas dinitrogen tetroxide decomposes to the brown coloured air pollutant nitrogen dioxide and exists in equilibrium. A 0.125 mol sample of dinitrogen tetroxide is introduced into a 1.00 L container and allowed to decompose at a given temperature. When equilibrium is reached, the concentration of the dinitrogen tetroxide is 0.0750 mol/L. What is the value of Keq for this reaction

$N^2{O}^4\rightleftharpoons 2N{O}^2$

0.0750 vs

	N2O4	2NO2
I	0.125	0
C	0.05	0.1
E	0.0750	0.1

$$\begin{array}{rl}& K_{eq}=\frac{[N{O}^2{]}^2}{[N^2{O}^4]}\\ & K_{eq}=\frac{0.1^2}{0.0750}\\ & K_{eq}=0.1333333333\dots \end{array}$$

11

At 100°C the reaction below has an equilibrium constant, Keq, value of 2.2x10-10. If 1.00 mol of phosgene, COCl2, is placed in a 10.0 L flask, calculate the concentration of carbon monoxide at equilibrium

$COC{l}_2(g)\rightleftharpoons CO(g)+C{l}_2(g)$

	COCL2	CO	CL2
I	0.1	0	0
C	-x	+x	+x
E	0.1-x	x	x

$$\begin{array}{rl}& K_{eq}=\frac{[CO][C{L}^2]}{[CoC{L}^2]}\\ & 2.2\ast 10^{-10}=\frac{[x{]}^2}{[0.1]}\\ & x^2=(2.2\times 10^{-10})\ast 0.100\\ & x^2=2.2\times 10^{-11}\\ & x=\sqrt{2.2\times 10^{-11}}\\ & x\approx 4.69E-6\end{array}$$

$\therefore$ the concentration is approx 4.69 x 10⁻⁶ M

2025-03-25

$Pb{l}_2(s)<->P{b}^2++2I^{-}$

$$\begin{array}{r}8.5\times 10^{-9}\end{array}$$ $$\begin{array}{rl}& K_{sp}=[P{b}^{2+}][2i^{-}]\\ & 8.5\ast 10^{-9}=[x][x]\\ & 8.5\ast 10^{-9}=x^2\\ & \sqrt{8.5\ast 10^{-9}}=\sqrt{x^2}\\ & 9.21954446e-5=x\end{array}$$

Lead KSP

\begin{align} & \frac{0.85g}{100 mL} \\ & M_{Pbl_{4}} = \frac{714.8g}{mol} \\ & 0.85g \to 0.001189 mol \\ & 100mL = 0.100 L \\ & \therefore \text{ the concentration at equilibrium is 1.2 * 10\^\;-2 mol/L} \end{align} $$\begin{array}{rl}& K_{sp}=[P{b}^{4+}][I^{-}{]}^4\\ & \dots solvethis\end{array}$$

Determine the ion concentrations of a saturated iron (II) hydroxide solution, where solubility is 0.72 g/100 mL. Solve to find the Ksp.

1 Write balanced chemical equations and the $K_{sp}$ for the following dissolving in water

a) Sodium Sulfide

$$\begin{array}{rl}& N{a}_{2}S(s)+H_{2}O(l)\to 2N{a}^{+}(aq)+S^{2-}(aq)\\ & K_{sp}=[N{a}^{+}{]}^2[S^{2-}]\end{array}$$

b) Calcium Iodide

$$\begin{array}{rl}& Ca{I}_2(s)+H_{2}O(l)\to C{a}^{2+}(aq)+2I^{-}(aq)\\ & K_{sp}=[C{a}^{2+}][I^{-}{]}^2\end{array}$$

c) Lithium Carbonate

$$\begin{array}{rl}& L{i}_{2}C{O}_3^{2-}(s)+H_{2}O(l)\to 2L{i}^{-}+C{O}_3^{2-}\\ & K_{sp}=[L{i}_2][C{O}_3^{2-}]\end{array}$$

d) Iron (II) sulfate

$$\begin{array}{rl}& FeS{O}_4^{2-}(s)+H_{2}O(l)\to F{e}^{2+}(aq)+S{O}_4^{2-}(aq)\\ & K_{sp}=[F{e}^{2+}][S{O}_4^{2-}]\end{array}$$

e) cobalt (II) nitrate

$$\begin{array}{rl}& Co(N{O}_3{)}_2(aq)+H_{2}O(l)\to C{o}^{2-}+2N{O}_3(aq)\\ & K_{sp}=\frac{[C{o}^{2-}][N{o}_3{]}^2}{[Co(N{O}_3{)}_2]}\end{array}$$

f) barium phosphate

$$\begin{array}{rl}& B{a}_3(P{O}_4{)}_2(s)+H_{2}O(l)\to 3B{a}^{+2}(aq)+2P{O}_4^{-3}(a)\\ & K_{sp}=[B{a}^{2+}{]}^3[P{O}_4^{-3}{]}^2\end{array}$$

5. What are the ion concentrations in saturated solutions of the following

Lead (II) Iodide TODO RECHECK THIS

$$\begin{array}{rl}& Pb{I}_2(s)\to P{b}^{2+}(aq)+2I^{-}(aq)\\ & K_{sp}=[P{b}^{+}][I^{-}{]}^2\\ & 9.8\times 10^{-9}=[x][x{]}^2\\ & \sqrt[3]{9.8\times 10^{-9}}=\sqrt[3]{x^3}\\ & 2.13997496e-9=x\\ & 2.14E-3=x\\ & \\ & \therefore P{b}^{2+}=2.14E-3andI=4.28E-3\end{array}$$

6 calculate the K sp values of the substances below from their solubility in water

a) thallium(l) chloride, 3.4g/L @ 25c

2025-03-26

$$\begin{array}{rl}& k_{eq}=8.5\ast 10^{-}12\end{array}$$

solve for x $(2x{)}^2(0.1+x)=8.5\ast 10^{-}12$ and $(2x{)}^2(0.1)=8.5\ast 10^{-}12$ todo

$27.8gPbC{l}_2\to \frac{0.1mol}{0.25L}PbC{l}_2\ast 4=0.4\frac{mol}{L}PbC{l}_2$
$PbC{l}_2\to P{b}^{2+}+$

Sample problem

425mL of 0.15M calcium Chloride
420mL of 0.25M Silver nitrate

Determine:

• a) If a precipitate will form
• b) The concetration of all ions at equilbirium
• c) the mass of the precipitate

$CaC{l}_2+2AgN{O}_3\to Ca(N{O}_3{)}_2+2AgCl$

1. figure out which product is least soluable (would precipitate out first)
   1. CaNo3 is soluable
   2. AgCl
2. 2. To see if a preipiate will form we need to calculate the Q and compare to the Ksp of AgC
   3. $AgC{l}_2<->A{g}^{+}+c{l}^{-}\to K_{sp}=[Ag][Cl]$
   4. Total volume = 425mL + 450Ml → 0.875 L
   5. mol Ag+ = C*V = 0.1125 mol
   6. mol Cl (425mL of 0.15M) * 2 = 0.1275mol
   7. [ag+] 0.1125 mol / 0.875 L= 0.12857
   8. [cl-] 0.1275 mol / 0.875 L = 0.145714

b)

• mol ca2+ 0.425L * 0.15M = 0.06375 mol
• $[C{a}^2+]$ = 0.06375 / 0.875 = … todo

c) AgCl is the precipitate
could total - could disolve = precipitated agcl

$=0.1125mol-(1.035\ast 10^{-8}mol\ast 0.875L$
$=0.11249999\dots mol$
$=0.1125$