Since the coefficients are only determined up to a multiplicative constant, set c 1 1 and solve for the coefficients: c 1 1 c c 3 1

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1 In[1]:= dipole moment of S4 Input interpretation: sulfur tetrafluoride dipole moment Result:.632 D (debyes) Unit conversions: pc m (picocoulomb meters) nc m (nanocoulomb meters) C m (coulomb meters).2486 au's of electric dipole moment In[2]:= C3O + O2 = CO2 + 2O Input interpretation: C 3 O (methanol) + O 2 (oxygen) CO 2 (carbon dioxide) + 2 O (water) Balanced equation: Balance the chemical equation: C 3 O + O 2 CO O Start over Add coefficients to all the molecules: c 1 C 3 O + c 2 O 2 c 3 CO 2 + c 4 2 O The number of C,, and O atoms on both sides of the reaction must be equal: C: c 1 c 3 : 4 c 1 2 c 4 O: c c 2 2 c 3 + c 4 Since the coefficients are only determined up to a multiplicative constant, set c 1 1 and solve for the coefficients: c 1 1 c c 3 1

2 2 c 3 1 c 4 2 Since one of the coefficients is a fraction, set c 1 2 and repeat the step above to obtain integer coefficients: c 1 2 c 2 3 c 3 2 c 4 4 Since the coefficients are all integers with a greatest common denominator equal to 1, substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: 2 C 3 O + 3 O 2 2 CO O Structures: Skeletal structure + O O O O O C + O Names: methanol + oxygen carbon dioxide + water Reaction thermodynamics: Enthalpy: Δ rxn kj/mol kj/mol = kj/mol (exothermic) Gibbs free energy: ΔG rxn kj/mol kj/mol = - 21 kj/mol (exergonic) Entropy: ΔS rxn 77.6 J/(mol K) J/(mol K) = J/(mol K) (exoentropic) Equilibrium constant: K c = [CO 2 ]2 [ 2 O] 4 [C 3 O] 2 [O 2 ] 3 Chemical names and formulas: methanol oxygen carbon dioxide water

3 3 formula C 3 O O 2 CO 2 2 O ill formula C 4 O O 2 CO 2 2 O name methanol oxygen carbon dioxide water IUPAC name methanol molecular oxygen carbon dioxide water Substance properties: methanol oxygen carbon dioxide water molar mass g/mol g/mol g/mol g/mol phase liquid (at STP) gas (at STP) gas (at STP) liquid (at STP) melting point -98 C -218 C C (at triple point) boiling point 64.7 C C C (at sublimation point) C C density.791 g/cm g/cm 3 (at C) g/cm 3 (at 2 C) 1 g/cm 3 solubility in water miscible surface tension.1347 N/m.728 N/m dynamic viscosity Pa s Pa s Pa s Pa s odor pungent odorless odorless odorless In[3]:= 5 grams N2 + 2 grams 2 = N3 Assuming 2 is a chemical compound Use as an isotope instead Input interpretation: 5 g of N 2 (nitrogen) + 2 g of 2 (hydrogen) N 3 (ammonia) Balanced equation: Step- by- step N N 3

4 4 Stoichiometry: ind the theoretical yield of the following reaction given 5 g N 2 and 2 g 2 : 5 g of N 2 (nitrogen) + 2 g of 2 (hydrogen) N 3 (ammonia) Start over Convert the specified mass of N 2 into moles using the molar mass ( g/mol). Convert the specified mass of 2 into moles using the molar mass ( g/mol): 1 5 g mol N g/mol g mol g/mol 2 In order to identify the limiting reactant, make a table of the molar quantities of the reagents corresponding to mol N 2 and mol 2 (one row for each). Begin by filling in these molar quantities: N 2 2 N mol N mol mol mol Write the balanced equation for the reaction: N N 3 Use the ratios of coefficients in the balanced equation to compute the molar quantities of the remaining reagents corresponding to mol N 2 : ( mol N 2 ) 3 mol 2 1 mol N mol 2 ( mol N 2 ) 2 mol N 3 1 mol N mol N 3 Summarize the results of the previous step in the table: N 2 2 N mol N mol mol mol mol mol Use the ratios of coefficients in the balanced equation to compute the molar quantities of the remaining reagents corresponding to mol 2 : ( mol 2 ) 1 mol N 2 3 mol mol N 2

5 5 ( mol 2 ) 2 mol N 3 3 mol mol N 3 Summarize the results of the previous step in the table: N 2 2 N mol N mol mol mol mol mol mol mol The limiting reactant (the row with the smallest values) is mol N 2 : N 2 2 N mol N mol mol mol mol mol mol mol The theoretical yield of the products will be: Answer: mol N 3 Structures: N N + N Skeletal structure Names: nitrogen + hydrogen ammonia Reaction thermodynamics: Enthalpy: Δ rxn kj/mol - kj/mol = kj/mol (exothermic) Gibbs free energy: ΔG rxn kj/mol - kj/mol = kj/mol (exergonic) Entropy: ΔS rxn 386 J/(molK) J/(molK) = -151 J/(molK) (exoentropic)

6 6 Equilibrium constant: K c = [N 3 ]2 [N 2 ] [ 2 ] 3 Chemical names and formulas: nitrogen hydrogen ammonia formula N 2 2 N 3 ill formula N N name nitrogen hydrogen ammonia IUPAC name molecular nitrogen molecular hydrogen ammonia Substance properties: nitrogen hydrogen ammonia molar mass g/mol g/mol g/mol phase gas (at STP) gas (at STP) gas (at STP) melting point -21 C C C boiling point C C C density.1251 g/cm 3 (at C) g/cm 3 (at C) g/cm 3 solubility in water insoluble surface tension.66 N/m.234 N/m dynamic viscosity Pa s Pa s Pa s odor odorless odorless WolframAlpha::nopst : The StoichiometryPod:ChemicalReactionData pod is not reporting any additional information for the StoichiometryPod:ChemicalReactionData Show all steps pod state. In[4]:= 5 grams N2 + 2 grams 2 = 4 grams N3 Assuming 2 is a chemical compound Use as an isotope instead

7 7 Input interpretation: 5 g of N 2 (nitrogen) + 2 g of 2 (hydrogen) 4 g of N 3 (ammonia) Balanced equation: Step- by- step N N 3 Stoichiometry: Show volume Step- by- step reagent N 2 2 N 3 name nitrogen hydrogen ammonia amount mol (5 g) mol (2 g) mol (4 g) equivalents 1 eq (limiting reactant) eq theoretical yield mol (6.794 g) % yield % amount remaining (none) mol (.9258 g) Structures: N N + N Skeletal structure Names: nitrogen + hydrogen ammonia Reaction thermodynamics: Enthalpy: Δ rxn kj/mol - kj/mol = kj/mol (exothermic) Gibbs free energy: ΔG rxn kj/mol - kj/mol = kj/mol (exergonic) Entropy: ΔS rxn 386 J/(molK) J/(molK) = -151 J/(molK) (exoentropic) Equilibrium constant: [N ] 2

8 8 K c = [N 3 ] [N 2 ] [ 2 ] 3 Chemical names and formulas: nitrogen hydrogen ammonia formula N 2 2 N 3 ill formula N N name nitrogen hydrogen ammonia IUPAC name molecular nitrogen molecular hydrogen ammonia Substance properties: nitrogen hydrogen ammonia molar mass g/mol g/mol g/mol phase gas (at STP) gas (at STP) gas (at STP) melting point -21 C C C boiling point C C C density.1251 g/cm 3 (at C) g/cm 3 (at C) g/cm 3 solubility in water insoluble surface tension.66 N/m.234 N/m dynamic viscosity Pa s Pa s Pa s odor odorless odorless In[5]:= S4 Assuming S4 is a chemical compound Use as a gene or a protein or more instead Use S as a unit instead Input interpretation: sulfur tetrafluoride

9 9 Chemical names and formulas: formula S 4 name sulfur tetrafluoride Lewis structure: Start over Draw the Lewis structure of sulfur tetrafluoride. Start by drawing the overall structure of the molecule: S Count the total valence electrons of the fluorine n,val 7 and sulfur n S,val 6 atoms: 4 n,val + n S,val 34 Calculate the number of electrons needed to completely fill the valence shells for fluorine n,full 8 and sulfur n S,full 8 : 4 n,full + n S,full 4 Subtracting these two numbers shows that bonding electrons are needed, which are already accounted for in the structure. Note that the valence shell of sulfur has been expanded. After accounting for the expanded valence, there are 4 bonds and hence 8 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain electrons left to draw: Answer: S Basic properties: molar mass phase g/mol gas (at STP)

10 1 melting point C boiling point 4.84 C density.4417 g/cm 3 solubility in water decomposes Gas properties (at STP): density.4417 g/cm 3 vapor density molar volume 3.78 (relative to air) cm 3 /mol Thermodynamic properties: specific free energy of formation Δ f G gas kj/g specific heat of formation Δ f gas kj/g critical temperature 364 K (at STP) Chemical identifiers: CAS number PubChem CID number Toxicity properties: short- term exposure limit 1 mg/m 3 link = information about chemistry features; link = how to access via mathematica

2) C 2 H 2 (g) + 2 H 2 (g) ---> C 2 H 6 (g) Information about the substances

2) C 2 H 2 (g) + 2 H 2 (g) ---> C 2 H 6 (g) Information about the substances Thermochemistry 1) 2 C 4 H 10 (g) + 13 O 2 (g) ------> 8 CO 2 (g) + 10 H 2 O(l) The reaction represented above is spontaneous at 25 C. Assume that all reactants and products are in their standard states.