{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2080_Prelim_I_review

# 2080_Prelim_I_review - 1 Three metals touch such that heat...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1.) Three metals touch such that heat transfer can occur between A and B or B and C, but no heat is directly transferred from C to A. i.) What is the final temperature of metals A, B, and C if: A has a specific heat of 0.30 J/g°C, mass of 15 g and an initial temperature of ‐8 °C B has a specific heat of 0.60 J/g°C, mass of 10.0 g and an initial temperature of 37 °C C has a specific heat of 0.30 J/g°C, mass of 4.0 g and an initial temperature of 23 °C? ii.) Does the answer to part i change if all three metals touch each other. 2.) The coil of a heating devise sits in an electrically insulating liquid A. When the heating devise is turned on for 5.00 seconds at 36 volts, the volume of gaseous A goes from 0.0 to 0.45 L. i.) ii.) If Pext = 1 atm what current (I) was used in the heating devise. Assume the volume and temperature (298 K) of the liquid do not change throughout the process. What is the ∆Hvaporization for the liquid under the temperature and pressure conditions given? 3.) The following ∆H° values are useful in finding ∆H°f for NaCl(s) from sodium metal and chlorine gas. However, one equation necessary for finding ∆H°f is missing. (a) Write out the missing chemical equation? (b) What is the ∆H° value for the missing chemical equation? (c) Write out the name of ∆H°x. i.e, standard enthalpy of ________________________________. 1/2Na+(g) + 1/2e‐ 1/2Na(g) ∆H°1 = ­248 kJ Na(s) Na(g) ∆H°sublimation = 108 kJ Cl(g) + e‐ Cl‐(g) ∆H°electron affinity = ­349 kJ Na+(g) + Cl‐(g) NaCl(s) ∆H°lattice formation = ­788kJ ___________________________________________________________ Na(s) + 1/2Cl2(g) NaCl(s) ∆H°f = ­411 kJ 4.) Consider the base catalyzed reaction at 298.15 K OCl‐ (aq) + I‐ (aq) OI‐ (aq) + Cl‐ (aq) a) Use the following initial‐rate data to determine the rate law and rate constant Rate = k[OCl‐]n[I‐]m[OH‐]p [OCl­] (M) 1.62 x 10‐3 1.62 x 10‐3 2.71 x 10‐3 1.62 x 10‐3 b) What is the overall order of this reaction? [I­] (M) 1.62 x 10‐3 2.88 x 10‐3 1.62 X 10‐3 2.88 x 10‐3 [OH­] (M) 0.52 0.52 0.84 0.91 Rate [M/s] 3.06 x 10‐4 5.44 x 10‐4 3.16 x 10‐4 3.11 x 10‐4 c) Suppose this reaction were run at 323.15 K and the new rate constant is calculated to be k = 606 s‐1. Calculate Ea and A for this reaction. 5.) The decomposition of dinitrogenpentoxide (N2O5) is a first order process: N2O5 (g) 2 NO2 (g) + ½ O2 (g) Rate = ‐d[N2O5]/dt = kobs[N2O5] At room temperature (298.15 K), the rate constant for this process is kobs = 3.5 x 10‐5 s‐1. A sample of 1.00 g of gaseous N2O5 is put into an evacuated 15.0 L flask at 298.15 K. The molar mass of N2O5 is 108.0 g/mole. Assume the temperature remains constant throughout the degradation. a) What is the initial partial pressure, in mmHg, of N2O5(g)? b) What is the partial pressure, in mmHg, of N2O5(g) after 3.0 hours? c) What is the total gas pressure, in mmHg, in the container after 3.0 hours? d) What is the half life (t1/2) of the reaction in hours? 6.) The gas‐phase reaction of chlorine with chloroform is described by the equation Cl2(g) + CHCl3(g) HCl(g) + CCl4(g) The rate law determined from experiment has a non‐integer order: Rate =k[Cl2]1/2[CHCl3] A proposed mechanism for this reaction follows: k1 Cl2 (g) ⇔ 2 Cl (g) Both fast with equal rates k‐ 1 e) Use the starting concentration of N2O5 (g) find the rate of decomposition of N2O5 (g). Using the reaction given, find the rate of formation of O2 (g) and NO2 (g). k2 Cl (g) + CHCl3 (g) HCl (g) + CCl3 (g) k3 CCl3 (g) + Cl (g) CCl4 (g) Slow Fast a) Is this an acceptable mechanism for the reaction? b) If the rate law was found to be rate = k [Cl2] what would the slow step be? ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online