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Edexcel Jan 2017 (IAL) Paper 4 Q24
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![Mark Partial pressures Round brackets in place of square brackets + symbol instead of multiply on top line Reject Mark Acceptable Answers Reject Kc = [NO]2[Cl2] [NOCl]2 IGNORE State symbols Acceptable Answers MARK CONSEQUENTIALLY ON EXPRESSION IN (i) NOCl NO(g) Cl2 Mol at start 2.00 0 0 Mol at eqm 1.780 (0.220) 0.110 (1) Concs /mol dm-3 0.356 0.044 0.022 (= mols at eqm 5) This may be shown as mols at eqm 5 in Kc expression (1) Kc = ((0.044)2 x (0.022)) = 3.36 x 10-4 mol dm-3 (0.356)2 Value IGNORE sf except 1sf (1) Units (1) Mark independently, consistent with Kc expression in (i) Correct final answer without working scores 4 marks Question Number 24a(i) Question Number *24a(ii)<br />
Acceptable Answers Kc is the same asEITHER Reject Mark temperature is unchanged OR it is unaffected by change is to volume / pressure / concentration (1) More NO (and Cl2) is formed because the quotient of the Kc expression decreases to keep Kc constant ALLOW More NO (and Cl2) forms because the pressure is reduced, so the reaction goes to the side with more (gas) moles OR More NO (and Cl2) forms because the pressure is reduced, so the reaction goes to the right (1) Mark independently Acceptable Answers Reject Hf So 298 NO +90.2 210.7 Cl2 0 (165.0) All three values (2) Any two values (1) Blank space or a dash instead of 0 Mark Question Number *24a(iii) Question Number 24b(i)<br />
Reject Mark Acceptable Answers Final answer of H = (+) 77(.0) kJ mol-1 scores 2 First mark : H = (2x90.2)(2x51.7) OR Hess cycle 2NOCl(g)2NO(g) + Cl2(g) (2x51.7) (2x90.2) N2(g)+O2(g)+Cl2(g) (1) H = (+) 77(.0) (kJ mol-1) (1) IGNORE Units ALLOW Max (1) TE for using a value other than 0 for Cl2 Acceptable Answers Reject Ssurroundings = - H/T ALLOW S = - H/T as long as there is reference to surroundings subsequently (1) (As H is positive), when T increases, Ssurroundings becomes less negative (so Stotal becomes less negative) IGNORE smaller and decreasing for less negative (1) No TE for MP2 if answer to (ii) is negative Acceptable Answers Reject S(total) = R lnK (1) IGNORE Kc / Kp K increases as T increases because EITHER S(total) increases (as T increases) OR Equilibrium moves to the right (as T increases) (1) Mark Mark Question Number 24b(ii) Question Number 24b(iii) Question Number 24b(iv)<br />
Mark Acceptable Answers Reject Answers discussing entropy change, not entropy 2nd mark dependent on 1st, for both methods. EITHER (Kinetic) energy of each particle is greater (1) ALLOW substances for particles So more ways of arranging particles or quanta / more disorder/ more random movement (at higher T) (1) IGNORE More collisions OR At the higher temperature the Maxwell- Boltzmann curve is more spread out (1) So there is greater randomness in the distribution of energies/ speeds (1) Question Number 24c(i)<br />
FOR 24c(ii) and 24c(iii): if mol-1 is written as mol-, only penalise once Question Number 24c(ii) Acceptable Answers Reject Mark Mark Reject +40.7 J mol-1 K-1 scores 2 marks Ssys = (189.3 + 2(231.2) -2(305.5)) (1) Magnitude, sign and units (1) No TE on incorrect expression ALLOW +63 J mol-1 K-1 for 1 mark due to using data at 298K Acceptable Answers Method 1 Ssurr = - H/T or use of expression e.g. -53.2 x 1000/800 (1) Value of Ssurr with sign and unit (-66.5 J mol-1 K-1 / -0.0665 kJ mol-1 K-1) OR Value of Stotal with sign and unit (-66.5 + 40.7 = -25.8 J mol-1 K-1 / -0.0258 kJ mol-1 K-1) (1) Stotal negative so not spontaneous ALLOW TE on incorrect S values in (ii) and (iii) If this gives a positive value for Stotal , then spontaneous (1) Method 2 When Stotal = 0, then T Ssystem = H (1) T= H/ Ssystem = 53200/40.7 = 1307 K (1) At T < 1307 K reaction is not spontaneous (1) Method 3 G = 53200 - 800 x 40.7 / = 53.2 - 800 x (40.7/1000) (1) = + 20640 J mol-1 / + 20.6 kJ mol-1 (1) G positive so reaction is not spontaneous (1) Question Number 24c(iii) (Total for Question 24 = 22 marks)<br />](https://learnah.org/wp-content/uploads/2018/08/January_2017_IAL_MS_-_Unit_4_Edexcel_Chemistry_A-level_24-0-1.jpg)
