Math  /  Data & Statistics

QuestionBEST Completer the statement or anwers the quetion. following reaction will Ke=KpK_{e}=K_{p} ? B) H2( g)+Br2( g( g)CO(g)+3H2( g)\mathrm{H}_{2(\mathrm{~g})}+\mathrm{Br}_{2}\left(\mathrm{~g}(\mathrm{~g}) \leftrightharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\right. C) N2O4( g)2 g)2HBr(g)\left.\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \leftrightharpoons 2 \mathrm{~g}\right) \leftrightharpoons 2 \mathrm{HBr}(\mathrm{g}) D) CO(g)+2NO2( g)\mathrm{CO}(\mathrm{g})+2 \mathrm{NO}_{2}(\mathrm{~g}) E) N2( g)+2H2( g)CH3OH(g)\mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g}) \leftrightharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})
7. Express the equilibrium constant for the following reaction. P4( g)+5O2( g)P4O10( s)\mathrm{P}_{4}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \leftrightharpoons \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{~s}) A) K=[P4][O2]5[P4O10]K=\frac{\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}}{\left[\mathrm{P}_{4} \mathrm{O}_{10}\right]} B) K=[P4O10][P4][O2]5K=\frac{\left[\mathrm{P}_{4} \mathrm{O}_{10}\right]}{\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}} C) K=[O2]5K=\left[\mathrm{O}_{2}\right]^{5} D) K=[O2]5K=\left[\mathrm{O}_{2}\right]^{-5} E) K=[P4O10][P4][O2]1/5K=\frac{\left[\mathrm{P}_{4} \mathrm{O}_{10}\right]}{\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{1 / 5}}
8. A solution is prepared by dissolving 0.23 mol of benzoic acid and 0.27 mol of potassium benzoate in sufficient water to yield 1.00 L of solution. The addition of 0.05 mol of HCl to this buffer soltation causes the pH to decrease only slighly. The pH does not decrease drastically because the HCl reacts with the \qquad present in the buffer solution. A) H2O\mathrm{H}_{2} \mathrm{O} B) H3O+\mathrm{H}_{3} \mathrm{O}^{+} O) porassium ion D) benzoic acíd E) benzoate ion
9. A 50.0 mL sample of an aqueous H2SO4\mathrm{H}_{2} \mathrm{SO}_{4} solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached upon addition of 125.1 mL of the base. The concentration of H2SO4\mathrm{H}_{2} \mathrm{SO}_{4} is \qquad MM. A) 0.234 B) 0.469 C) 0.150 D) 0.300 E) 0.938 Page 2 of 6

Studdy Solution

STEP 1

What is this asking? We need to find which reaction has Ke=KpK_e = K_p, write the equilibrium constant for a reaction involving a solid, figure out why adding HCl to a buffer doesn't change the pH much, and calculate the concentration of sulfuric acid given titration data. Watch out! Remember solids don't appear in equilibrium expressions, and strong acids react completely with conjugate bases in buffer solutions.
Don't forget the stoichiometry when dealing with acids and bases!

STEP 2

1. Find the reaction where Ke=KpK_e = K_p
2. Write the equilibrium constant expression
3. Explain the buffer behavior
4. Calculate the sulfuric acid concentration

STEP 3

Remember, the relationship between KeK_e and KpK_p is given by Kp=Ke(RT)ΔnK_p = K_e(RT)^{\Delta n}, where Δn\Delta n is the change in the number of moles of gas.

STEP 4

We're looking for a reaction where Ke=KpK_e = K_p.
This happens when Δn=0\Delta n = 0.
Let's check each reaction: * B) Δn\Delta n = 4 - 2 = 2.
Nope! * C) Δn\Delta n = 2 - 2 = 0.
Bingo! This is it! * D) Δn\Delta n = 1 - 3 = -2.
Nope! * E) Δn\Delta n = 1 - 3 = -2.
Nope!

STEP 5

So, the answer is C, where Δn=0\Delta n = 0, making Kp=KeK_p = K_e.

STEP 6

Remember, solids and pure liquids don't show up in equilibrium constant expressions!

STEP 7

For the reaction P4(g)+5O2( g)P4O10(s)\mathrm{P}_{4}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{~g}) \leftrightharpoons \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{s}), since P4O10\mathrm{P}_{4} \mathrm{O}_{10} is a solid, it won't be included in our expression.
So, K=1[P4][O2]5K = \frac{1}{\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}}.

STEP 8

This matches answer D, if we rewrite our expression as K=[O2]5K = \left[\mathrm{O}_{2}\right]^{-5}.

STEP 9

Buffers resist pH changes because they contain a weak acid and its conjugate base (or a weak base and its conjugate acid).

STEP 10

When we add HCl, a strong acid, it reacts completely with the **benzoate ion**, the conjugate base.
This reaction consumes the added H+H^+ ions, preventing a drastic pH change.

STEP 11

The answer is **E**, benzoate ion.

STEP 12

The balanced reaction is: H2SO4+2NaOHNa2SO4+2H2O\mathrm{H}_{2} \mathrm{SO}_{4} + 2\mathrm{NaOH} \leftrightharpoons \mathrm{Na}_{2} \mathrm{SO}_{4} + 2\mathrm{H}_{2} \mathrm{O}.

STEP 13

We have 125.1 mL125.1 \ \mathrm{mL} of 0.375 M0.375 \ \mathrm{M} NaOH, so the number of moles of NaOH is (125.1 mL)(0.375 mmol/mL)=46.9125 mmol(125.1 \ \mathrm{mL}) \cdot (0.375 \ \mathrm{mmol/mL}) = 46.9125 \ \mathrm{mmol}.

STEP 14

From the balanced equation, 22 moles of NaOH react with 11 mole of H2SO4\mathrm{H}_{2} \mathrm{SO}_{4}.
So, the moles of H2SO4\mathrm{H}_{2} \mathrm{SO}_{4} are 46.9125 mmol2=23.45625 mmol\frac{46.9125 \ \mathrm{mmol}}{2} = 23.45625 \ \mathrm{mmol}.

STEP 15

The concentration of H2SO4\mathrm{H}_{2} \mathrm{SO}_{4} is 23.45625 mmol50.0 mL=0.469 M\frac{23.45625 \ \mathrm{mmol}}{50.0 \ \mathrm{mL}} = 0.469 \ \mathrm{M}.

STEP 16

The answer is **B**, 0.469 M.

STEP 17

The answers are: C, D, E, and B.

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