Extra Credit 82

Q2.67

Calculate the values of \(c_{rms}\), \(c_{mp}\), and \( c_{average}\) for \( O_2 \) at \(T = 298 K \)

Relevant Information:R = 8.314 J mol-1 K-1 and M = 0.032 kg mol-1

Q9.6

1. Calculate the rate constant for the reaction of the first-order decay of radioactive 32P, which has a half-life of about 14.3 days.
2. Calculate the age of an object, found by an archeology student while out in the field, using carbon dating. The object is observed to have a 14C content of 0.93x10-14 mol%. The natural abundance found in living matter is a 14C content of 1.1x10-3 mol%.

Q9.30

Determine the activation energy for a first order reaction with a specific rate constant of 8.76x10-5 s-1 at 303 K and 2.77x102 s-1 at 620 K.

Q10.5

An enzyme is assayed with an initial concentration of \( 5.6x10^{-5}M \). After \( 10 minutes \) when half of the enzyme is consumed, calculate Vmax.

Relevant information:

Km for the substrate is \( 7.98x10^{-5} M \) .

Q11.20

What affect would the emission of \( H_{2} \) show while taking an emission spectrum of hydrogen in a Balmer series?

Q12.14

Which of the following species has the shortest bond length: CO, CO+, CO2+?

Q13.15

Which of the following compounds is a stronger base:

\( CH_{3}NH_{2} \) or \( (CH_{3})_{2}NH \)

Q14.18

Give the number of normal vibrational modes in each of the following:

1. (a) \( H_{2} O \)
2. (b) \( CO_2 \)
3. (c) \( CH_{3} CCl \)
4. (d) \( SO_2 \)

Q2.74

There is a sample of an ideal gas in a sealed container at \( 295 K \) which is then heated to \(407 K \) with a final pressure of \( 2.85 atm \) . What would the initial pressure have been?

Q9.13

In a first order decomposition reaction, half of the compound decomposes in \( 20 minutes \). What is the rate constant of the reaction?

Answer Key

Q2.67: \( c_{rms}= 475 \; m/s \) , \( c_{mp} = 394 \; m/s \) , \( c = \; 444 m/s \)

Q9.6: (a) 4.8x10-2 days-1 (b) 2.03x104 years

Q9.30: Ea = 73.74 kJ

Q10.5: Vmax = 1.078x10-5 M/min

Q11.20: A spectral line in the Balmer series that is inconsistent with the Bohr theory

Q12.14: Shortest bond length: CO

Q13.15: (CH3)2NH

Q14.18: H2O: 3, CO2: 4, CH3CCl3: 18, SO2: 3

Q2.74: P = 2.07 atm

Q9.13: k = 0.0347 min-1

S2.67

Relevant information: \( T = 298 K \), R = 8.314J mol-1K-1, and M = 0.032 kg mol-1

The root mean square speed is given by the following equation:

\[ c_{rms}= \sqrt{ \dfrac{3RT}{M} }\]

\[ c_{rms}= \sqrt{ \dfrac{(3)(8.3145)(298\,K)}{0.032 \;kg/mol}} = 482 {\frac{m}{s}}\]

The most probable speed is found using a derivation of the Maxwell equation. The final step of the derivation is given by the equation:

\[v_{mp} = \sqrt{\frac{2k_bT}{m}} = \sqrt{\frac{2RT}{M}} = c_{mp}\]

\[c_{mp} = \sqrt{\frac{2RT}{M}} = \sqrt{\frac{2(8.3145 \;J \; mol^{-1} \; K^{-1})(298\;K)}{0.032 \;kg/mol}}= 393.5 {\frac{m}{s}}\]

The average speed is given by:

\[c_{average}= \sqrt {\dfrac{8RT}{\pi M}} \]

\[ c_{average} = \sqrt {\frac{8(8.3145 \;J \; mol^{-1} \; K^{-1})(298\;K)}{\pi{(0.032 \;kg/mol)}}} = 444.0{\frac{m}{s}}\]

S9.6

(a) To calculate the rate constant, you must use the half-life equation for first order reactions:

\[t_{1/2} = \dfrac{\ln2}{k}\]

\[14.3\ days = \dfrac{\ln2}{k}\]

Rearrange to get k:

\[k = \dfrac{\ln2}{14.3 days}\]

\[ k = 4.8X10^{-2}days^{-1} \]

(b) Important information: \( k = 1.212X10^{-4} years^{-1} \)

\[ t_{1/2} = -\dfrac{ln\dfrac{14C}{14C_o}}{k} \]

\[t_{1/2} = -\dfrac{1}{1.212X10^{-4}} \times{ln \dfrac{0.93X10^{-14}}{1.1X10^{-13}}} \]

\[t_{1/2}=2.03x10_{4} years \]

assuming that the natural abundance of 14C has not changed since the objects time period.

S9.30

To determine activation energy \( Ea \) from two temperatures and their corresponding rate constants, you must use the Arrhenius equation:

\[(\ln \dfrac{k_{2}}{k_{1}} = \dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right ))\]

\[(\ln \dfrac{277s^{-1}}{8.76x10^{-5}s^{-1}} = -\dfrac{912,000J}{8.314} \left (\dfrac{1}{620K}-\dfrac{1}{303K} \right ))\]

\[ Ea = 73.74kJ \]

S10.5

The equation for rate is given by:

\[ v_0 = \dfrac{V_{max} [S]}{K_M + [S]} \]

First we must calculate \(v_o\) by using the change in substrate concentrations divided by the change in time. We know that half of it is consumed in 10 minutes, so the change in substrate is \( = 2.8X10^{-5} \)

\[ v_o = \dfrac{2.8X10^{-5}}{10min} \]

\[v_o = 2.8X10^{-6} \]

Then you can rearrange the equation to solve for \(v_{max} \)

\[ V_{max} = v_0 \left ( \dfrac{K_M}{[S]} + 1 \right ) \]

\[ V_{max} = 2.8X10^{-6} \left ( \dfrac{7.98X10^{-5}}{2.8X10^{-5}} + 1 \right ) \]

\[V_{max} = 1.078X10^{-5} \; M/min \]

S11.20

If the gas is assumed to be pure, a hydrogen spectrum will show specific lines that are consistent with the Bohr model. This is due to the quantization of energy into different levels. The Balmer series can be used to describe a jump from a higher to a lower energy, and would create specific hydrogen emission spectra that are inconsistent with the Bohr emission lines.

S12.14

Bond length is inversely proportional to bond order. As bond order increases, bond length gets smaller. The equation for bond length is the following:

\[\text{Bond Order} = \dfrac{\text{(electrons in bonding orbitals - electrons in anti bonding orbitals)}}{2} \]

The number of electrons in the bond and antibonded orbitals can be determined through the electronic configurations.

CO:

\[ \left ( \sigma_{2s} \right ) ^{2}\left ( \sigma^{\star }_{2s} \right )^{2} \left ( \pi_{2p_{x,y}} \right )^{4} \left ( \sigma_{2p_{z}} \right )^{2} \]

\[\text{Bond Order}_{CO} = \dfrac{1}{2}\times(8 - 2) = 3\]

CO+:

\[ \left ( \sigma_{2s} \right ) ^{2}\left ( \sigma^{\star }_{2s} \right )^{2} \left ( \pi_{2p_{x,y}} \right )^{4} \left ( \sigma_{2p_{z}} \right )^{1} \]

\[\text{Bond Order}_{CO^{+}} = \dfrac{1}{2}\times(7 - 2) = 2.5\]

CO2+:

\[ \left ( \sigma_{2s} \right ) ^{2}\left ( \sigma^{\star }_{2s} \right )^{2} \left ( \pi_{2p_{x,y}} \right )^{4} \left ( \sigma_{2p_{z}} \right )^{0} \]

\[\text{Bond Order}_{CO^{2+}} = \dfrac{1}{2}\times(6 - 2) = 2\]

Bond order is inversely proportional to bond length so:

Shortest bond is \( CO \)

CO< CO+< CO2+

S13.15

The stronger base is (CH3)2NH when compared to CH3NH2 because it is a better H+ acceptor.

S14.18

To calculate number of vibrational modes, there are two different equations you can use. For nonlinear molecules we use the equation \(3N-6 \) for the normal modes where N is the number of atoms in the molecule. And for linear molecules we use \( 3N-5. \)

• H2O is non-linear --> \( 3 (3) - 6 = 3 \)
• CO2 is linear --> \( 3 (3) - 5 = 4 \)
• CH3CCl3 is non-linear --> \( 3 (8) - 6 = 18 \)
• SO2 is non-linear --> \( 3 (3) - 6 = 3 \)

S2.74

This problem can be solved using the ideal gas law:

\[ PV = nRT \]

Assuming P and T are the only things changing, while volume and number of moles are held constant, we can make the following ratio:

\[ \dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \]

Plugging in the values, we get:

\[ \dfrac{P_1}{295K} = \dfrac{2.85atm}{407K} \]

Solving for \( P_1 \) we get:

\[ P = 2.07 atm \]

S9.13

To determine the rate constant, you must use the equation for half life of a first order reaction.

\[t_{1/2} = \dfrac{\ln2}{k}\]

\[20\ minutes = \dfrac{\ln2}{k}\]

Rearrange to get k:

\[k = \dfrac{\ln2}{20\ minutes}\]

\[ k = 0.0347 \min^{-1} \]