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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf q2,4,6,7,810,11 these are the starting queshtions thank u actually
i'll answer them again :
q2 : there should be a distribution of energies starting frm zero onwards ...and an increse in temp shifts the molecular energis to the right but area under the two curves should be same since ther no. of molecules remains the same
q:4
use equation of bonds broken - bonds formed .....draw a displayed formula to get the picture clearly
q6: activation energy is between E1 and E2 and since E1 is greater you sbtract the two to get the humped area hieght (the transition state)
q7: this one you have to use ( answer by 6Astarstudent) Here is a quick equation
(V1P1 + V2P2)/(V1+V2) = new P
so (5x12 + 10x6)/(5+10) = 120/15 = 8
q8: draw an enthalpy cycle for easy calculation ....take the vlaue for Ionisation energies of Ca from the booklet
q10nly graphit is ringed structure here and does not have the tetrahedral structure of the other
q11: write intial moles and equlibrium moles ....multiple the 0.25 of y by 2 since mole ratio is 1:2
to calculate kc (0.25 x 2 )2 / 0.25 = 1