Calorimetry
For the following reaction:
NO2 (g) + CO (g) -> NO (g) + CO2 (g),
the rate law is: Rate = k[NO2]2. If a small amount of gaseous carbon monoxide (CO) is added to a reaction mixture that was 0.10 molar in NO2 and 0.20 molar in CO, which of the following statements is true?
a. Both k and the reaction rate remain the same b. Both k and the reaction rate increase
c. Both k and the reaction rate decrease
d. Only k increases, the reaction rate will remain the same
a) Both k and the reaction rate remain the same The value k remains the same unless the temperature is changed or a catalyst is added. Only materials that appear in the rate law (NO2) will affect the rate. Adding NO2 would increase the rate while moving it would decrease the rate. CO has no effect on the rate.
Calculate ΔH and ΔS for the following reaction in order to decide if ΔH or ΔS drives the reaction.
Delta H = -92.4 kJ/mol favorable
Delta S = -198.24 unfavorable
Delta G = -33.3 SPONTANEOUS so ΔH drives the reaction.
Calculate the entropy change at 25°C, in J/K for: 2SO2(g) + O2(g) → 2 SO3(g)
So = So(products) - So(reactants) = [2 mol SO33 x 256.6 J/mol-K] - [2 mol SO2 x 248.1 J/mol-K + 1 mol O2 x 205.3 J/mol-K] = -188.3J/K
All factors affect the rates of chemical reaction EXCEPT: a. Temperature
b. Catalyst
c. Concentration of reactants
d. Number of reactants
e. Physical state of reactants
d) Number of reactants The temperature, concentration of reactants, physical state of reactants, and presence of a catalyst all affect the rates of chemical reactions.
Calculate the standard enthalpy of combustion of the transition of C(s, graphite) → C(s, diamond),
C(s, graphite) + O2 → CO2 ΔHo = -393.5 kJ/mol CO2 → C(s, diamond) + O2 ΔHo = + 395.41 kJ/mol
ΔH overall = =1.9 kJ/mol
Calculate ΔH for this reaction: CH4(g) + NH3(g) --> HCN(g) + 3H2(g) given:
N2(g) + 3 H2(g) ---> 2 NH3(g) ΔH = -91.8 kJ
C(s)+ 2 H2(g) ---> CH4(g) ΔH = -74.9 kJ
H2(g)+2 C(s) +N2(g) ->2 HCN(g) ΔH= +270.3kJ
1) Analyze what must happen to each equation: a) first eq ⇒ flip and divide by 2 (puts one NH3 on the reactant side) b) second eq ⇒ flip (puts one CH4 on the reactant side) c) third eq ⇒ divide by 2 (puts one HCN on the product side) 2) rewite all equations with the changes: NH3(g) ---> (1/2)N2(g) + (3/2)H2(g) ΔH = +45.9 kJ <--- note sign change & divide by 2 CH4(g) ---> C(s) + 2 H2(g) ΔH = +74.9 kJ <--- note sign change (1/2)H2(g) + C(s) + (1/2)N2(g) ---> HCN(g) ΔH = +135.15 kJ <--- note divided by 2 3) What cancels when you add the equations: (1/2)N2(g) ⇒ first and third equations C(s) ⇒ second and third equations (1/2)H2(g) on the left side of the third equation cancels out (1/2)H2(g) on the right, leaving a total of 3H2(g) on the right (which is what we want) 4) Calculate the ΔH for our reaction: +45.9 kJ plus +74.9 kJ plus +135.15 = 255.95 kJ = 260. kJ (to three sig figs)
The integrated rate law for second order.
[A]-1t - [A]-10 = -kt
Consider the following reaction:
4 NH3(g) + 5 O2(g) ⇌ 6 H2O(g) + 4 NO(g)
Using thermodynamic data, decide if this reaction is spontaneous at 298 K.
ΔG⁰ = -958.5 Spontaneous
Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equations:
C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = -1299.5 kJ
C(s) + O2(g) ---> CO2(g) ΔH° = -393.5 kJ
H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ
1) Determine what we must do to the three given equations to get our target equation: a) first eq: flip it so as to put C2H2 on the product side b) second eq: multiply it by two to get 2C c) third eq: do nothing. We need one H2 on the reactant side and that's what we have. 2) Rewrite all three equations with changes applied: 2CO2(g) + H2O(ℓ) ---> C2H2(g) + (5/2)O2(g) ΔH° = +1299.5 kJ 2C(s) + 2O2(g) ---> 2CO2(g) ΔH° = -787 kJ H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ Notice that the ΔH values changed as well. 3) Examine what cancels: 2CO2 ⇒ first & second equation H2O ⇒ first & third equation (5/2)O2 ⇒ first & sum of second and third equation 4) Add up ΔH values for our answer: +1299.5 kJ + (-787 kJ) + (-285.8 kJ) = +226.7 kJ
The following mechanism has been proposed for the reaction of CHCl3 with Cl2.
Step 1: Cl2 (g) -> 2 Cl (g) fast
Step 2: + CHCl3 (g) -> CCl3 (g) + HCl (g) slow
Step 3: CCl3 (g) + Cl (g) -> CCl4 (g) fast
Which of the following rate laws is consistent with this mechanism?
Rate = k[Cl2]
Rate = k[CHCl3][Cl2]
Rate = k[CHCl3]
Rate = k[CHCl3][Cl2]1/2
d) Rate = k[CHCl3][Cl2]1/2 The rate law depends on the slow step of the mechanism. The reactants in the slow step are Cl and CHCl3 (one of each). The rate law is first order with respect to each reactant. The Cl is half of the original reactant molecule Cl2, so the [Cl] in the rate law is replaced with [Cl2]1/2.
The steps below represent a proposed mechanism for the catalyzed oxidation of CO by O3.
Step 1: NO2 (g) + CO (g) -> NO (g) + CO2 (g) Step 2: NO (g) + O3 (g) -> NO2 (g) + O2 (g)
What are the overall products of the catalyzed reaction? a. CO2 and O2 b. NO and CO2 c. NO2 and O2 d. NO and O2
a) CO2 and O2 Add the two equations together: NO2 (g) + CO (g) + NO (g) + O3 (g) -> NO (g) + CO2 (g) + NO2 (g) + O2 (g) Then cancel out identical species that appear on opposite sides: CO (g) + O3 (g) -> CO2 (g) + O2 (g)