Flashcards
Topic 11
Cards in this deck (57)
- 1How do we express equilibrium constants for reactions involving gases?expressed in terms of partial pressures
- 2what is partial pressure?-in a mixture of gases the partial pressure of a gas is the pressure that the gas would exert on its own under the same conditions
- 3give the equation for calculating partial pressurePA = XAPt
Pa= partial pressure of gas A
Xa= mole fraction of gas A
PT= total pressure - 4give the equation for calculating mole fraction of gas AXa = amount of A divided by
total amount of gas - 5why does mole fractions have no units?-as the amounts of each gas are measured in the same units (mol)
- 6Hydrogen iodide, hydrogen + iodine may reach equilibrium: 2HI (g) = H2 (g) + I2 (g) 4.0 mol of HI was sealed in a flask at 2.0 atm pressure + heated. At equilibrium, there was 3.6 mol of HI (a) Calculate the equilibrium amount of H2Amount of HI reacted = 4.0- 3.6 = 0.4 mol
2 mol of HI produce 1 mol of H2 hence amount of HI reacted is divided by 2
Amount of H2 produced = 0.4/2 = 0.2 mol - 7(b) calculate the partial pressure of hydrogenAmount of I2 = amount of H2= 0.2 mol
Total amount of gas = 3.6 + 0.2 +0.2
= 4.0 mol
Mole fraction of H2 = 0.2/ 4.0 - 0.05
Partial pressure = 0.05 x 2.0 = 0.1 atm - 8Nitrogen reacts with hydrogen to form ammonia N2 (g) + 3H2 (g) = 2NH3 (g) write an expression for the equilibrium constant KpKp = (PNH3) ^2 / (PN2) (PH2) ^3
- 9Calcium carbonate decomposes when heated CaCO3 (s) = CaO (s) + Co2 (g) write an expression for the equilibrium constant, KpKp = PC02
(solids, liquids don't appear in an expression for Kp, only gases do) - 10what can you calculate if you have the equilibrium concentration of partial pressure?value of Kc , Kp
- 11how do you work out the units of Kc1) Write the expression for Kc
2) total the powers for the top conc's
3) total the powers for the bottom conc's
4) do (2) - (3)
5) apply answer 4 the concentration units, moldm-3 - 12work out the Kc for N2 (g) + 3H2 (g) = 2NH3 (g)Kc = [NH3 (g) ] ^ 2 / [N2 (g) ] [H2 (g) ] ^3
2-4 = -2
units of Kc are mol-2dm6
as applied to units moldm-3 e.g.
-2 x 1 = -2
-2 x -3 = 6 - 13explain how to work out the units for Kp1) write expression for Kp
2) total powers for top partial pressures
3) total powers for bottom partial pressures
4) (2) - (3)
5) apply answer 4 to pressure units, atm - 14work out Kp for N2 (g) + 3H2 (g) = 2NH3 (g)Kp = (PNH3) ^ 2 / (PN2)(PH2) ^3
2-4= -2
units for kp are atm-2 - 15when will Kc or Kp have no units ?if there are equal no. of particles on both sides of the equation
-point this out in an answer - 16N204 + N02 may reach equilibrium: N204 (g) = 2NO2 (g) 1.00 mol of N204 was sealed in a 0.5 dm3 flask. At equilibrium with N02, there was 0.6 mol of N204. Calculate the equilibrium concentrations of N204 + N02[N204] = 0.6/ 0.5 = 1.2 moldm-3
Amount of N204 reacted = 1.0 - 0.6
=0.4 mol
Amount of NO2 produced = 0.4 x 2
= 0.80 mol
[NO2] = 0.8/ 0.5 = 1.6 moldm-3 - 17Kc= [NO2 (g) ] ^ 2 / [N2O4 (g) ] Calculate the value of Kc, including its units refer to previous answerKc = 1.6 ^ 2 / 1.2
= 2.1
units = 2-1= 1
overall= 2.1moldm-3 - 18when can you ignore the volume for a Kc calculation?- if the value of Kc has no units
- 19How can the values of Kc or Kp be affected?- by changes in temp
- 20if you have a reversible reaction that is exothermic in the forward direction what happens to Kc/ Kp if temp increases?Kc/Kp decreases
as the product value decreases and reactant value increases
hence Kp= smaller value/ greater value
due to the position of equilibrium moving to the left - 21for a reversible reaction that is endothermic in the forward direction what happens to Kc/Kp with increasing tempKc/Kp increases
as product value increases + reactant value decreases
Kp = higher value/ lower value
- as position of equilibrium moves to the right - 22what effect does changing concentration of reacting substance have on Kc?- has no affect on Kc
read example of pg 79 for reason why - 23what is Qc?Qc is the expression for the reaction quotient of a reaction
same as expression for Kc but conc's not necessarily the ones at equilibrium - 24hydrogen can be manufactured by reacting carbon with steam C (s) + H20 (g) = CO (g) + H2 (g) Kp = (PCO)(PH2) / (PH20) explain why an increase in pressure leads to a decreases in the yield of hydrogenAs Kp is kept constant
when total pressure increased
mole fractions of products, Co + H2 must decrease will mole fraction of reactant H20 must increase
causing a reduce in yield of H2 - 25what affect does adding a catalyst have to value of Kc or Kp?-don't affect position of equilibrium or value of Kc, Kp
- but does increase the rates of the forward + backward reactions by the same ratio
-reduce time take to reach equilibrium - 26A buffer solution containing ethanoic acid + sodium ethanoate, is required with a pH of 5.06 a) calculate its H+ conc[H+]= 10^-pH
= 10^-5.06
=8.71 x 10 ^-6 moldm-3 - 27A buffer solution containing ethanoic acid + sodium ethanoate, is required with a pH of 5.06 b) Calculate the ration [CHCOOH]/[CH3COONa] required. (Ka = 1.74 x 10 ^ -5 moldm-3)[H+] = Ka x [acid]/[salt]
[CH3COOH]/[CH3COONa] = [H+]/Ka
= 8.71x10 ^ -6 / 1.74 x 10 ^-5
= 0.50 - 28A buffer solution containing ethanoic acid + sodium ethanoate, is required with a pH of 5.06 c) calculate the ratio [ CH3COOH]/[CH3COONa] pKa = 4.76[CH3COOH]/[CH3COONa] = 10^ (pKa- pH)
= 10^ (4.76-5.06)
=0.50 - 29A buffer solution containing ethanoic acid + sodium ethanoate, is required with a pH of 5.06 c) 0.2 moldm-3 ethanoic acid is supplied. Calculate the conc of sodium ethanoate solution required when equal volumes are mixed[CH3COONa] = 0.2 / 0.50
=0.4 moldm-3 - 30what happens to the Ph of strong acids as you dilute them?- the Ph increases by 1 for each x10 dilution of a strong monobasic acid
e.g. 1.0 moldm-3 of HCL = ph of 0 at 298K
0.1 moldm-3 of HCL = 1 = dilution factor x10
0.01 moldm-3 of HCL = 2 = factor x100 - 31what happens to the Ph when you dilute a weak acid?Ph increases by 0.5 for each x10 dilution of a weak monobasic acid
e.g. 1.0 moldm-3 ethanoic acid = ph of 2.38
at 0.1 = ph of 2.88= dilution factor x10
at 0.01= ph of 3.38 = dilution factor x100 - 32how do you achieve a x10 dilution of an acid?-need to mix...
a) 1 volume of the original solution
b) 9 volumes of water
e.g. 50cm3 of 1.0 mol HCL + 450cm of H20 = 500cm of 0.1 moldm-3 HCL - 33what factors affect the pH of an acidic buffer?-value of Ka (for weak acid)
-ratio of [Acid]/[salt] - 34does the value of Ka + ratio of [acid]/[salt] + pH of buffer solution stay the same when a buffer is diluted?- yes stay the same
-unless diluted to a large extent Ph of buffer stays the same when diluted - 35what does a titration curve show?-show how the pH changes when monobasic acids + bases are mixed
- 36draw a rough sketch of the titration curve of 0.1 moldm-3 HCL being added to 25cm3 of 0.1 moldm-3 NaOH in a conical flask
- 37what is the equivalence point?-point at which the acid + base been mixed in the exact proportions shown in the chemical equation
-for monobasic acids/bases this is when equimolar (same no. of moles of acid + base) have been mixed - 38identify where the equivalence point would be on a titration curvewhen pH halfway through the vertical section is achieved
e.g. if ph mixtures ranges up to 14, equivalence point = 7 - 39what is the end point in a titration?-point in the titration where the indicator changes colour
- both the equivalence + end point aren't the same as the "neutral point"= not always at pH 7 - 40describe the titration curve, concerning pH, when a strong acid is mixed into a weak base- ph falls rapids at start
-gradient of decline then reduces until close to the equivalence point due to a buffer solution forming - 41roughly sketch the titration curve of a strong acid being mixed into a weak base
- 42describe the titration curve, concerning pH, when a weak acid is mixed into a weak base- may be no obvious change in pH at equivalence point
-instead might be a point of inflexion with a slight change in gradient - 43draw a titration curve of a weak acid being mixed into a weak base
- 44sketch a titration curve for a weak acid added into a strong base. Indicate the pH at the start, equivalence, and twice equivalence
- 45concerning the curve just drawn.. what must you assume to obtain the answer?-assumes that 25cm3 of strong base is contained in the flask
- that acid + base are equal concentrations - 46notice after equivalence the gradient reduces, why is this?-because a buffer solution forms
consisting of a weak acid + salt of its conjugate base (formed during the titration) - 47how can you work out the pKa from a titration curve?for stong base- weak acid titrations
ph at half equivalence = pKa of acid - 48what is an acid base indicator?- usually weak acid
- has different colours depending on its pH - 49give the colour of methyl orange in acidic solutions, in alkaline solutionsacidic= red
alkaline = yellow - 50what determines the pH at which the indicator changes colour? explain this using methyl orange equation HIn (aq) = H+ (aq) + In- (aq)-determined by equilibrium constant Kin
Ki = [H+ (aq) ] [In- (aq)] / [HIn (aq)]
when [Hln] = [In-] methyl orange appears orange (as mix of red + yellow) - 51Kin = [H+ (aq)] [ In- (aq)] / [HIn(aq)] if you know [H+(aq)] = 2.00x 10-4 moldm-3 calculate the pH at which methyl orange changes colourKin = [H+ (aq)] = 2.00 x10-4
as other conc the same so cancel each other out
pKa= -log10 (2.00 x 10-4) = 3.70
Ph at which colour changes = 3.70 - 52what are single indicators? give 2 examples-provide a sharp end point
- methyl orange, phenolphthalein - 53list the situations when methyl orange is a suitable indicatorwhen a ... is present
- strong acid
-strong base + strong acid
-weak base - 54list the situations when phenolphthalein is a suitable indicatorwhen a....is present
-strong acid
-strong base + weak acid
-strong base - 55read through worked example pg 86 very important don't miss!!.....
- 56explain what is happening in this titration curve-ph initially rises rapidly due to a base being added to the weak acid
- gradient decreases because a buffer solution forms , which resists the changes in Ph
- pH rises rapidly near equivalence point
-ph rises slowly again because base is added to solution in which all the acid has now been neutralised
- 57—
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- GCSE Chemistry (Topic 10-14 ) Flashcards on Topic 11, created by Jade B on 29/03/2018.
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