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Help me in J5, Q2(b)(iii) and 3 (b)

hey im so sorry family probs i couldnt come at exact 11 im so sorry

its ok, its just 15 minutes up...

sure
q2 b(iii)
Mechanism B, because the rate-determining step is the slow step which should involve only (CH3)2Cdouble bond O and CN^-1 not H+ as H+ doesnt appear in the rate equation so must be involved after the slow step

btw check out page 329 from textbook as well
i hope that helps if not do text me

actually im very particualar about time im really sorry but next time i will make sure of no problems isA

3B is as follows

delts H= 82 - 178= -96 KJmol^-1 as 82 must be positive as the equation is shifted
i hope that helps

come on dear! I didn't take that seriously as u r taking it. I just turned my laptop on put near me. It didn't waste my time....haha..calm down. its okay. Dont stress urself..Exams are looming up..!!

oh thank u im sorry my brother got me in a bad mood sorry

but u know what. in 3b, i can't understand the formula or what ...why are we subtracting 178 from 82? I tried to construct the Hess cycle but could not do that

oh its okay i will help u understand it sadia sweeto
okay honey look, the 1st is CaCO3 arrow CaO + CO2 which is +178
and the 2nd one is CaO + H2O arrow Ca(OH)2 which is -82

the final is a combination of both thats all no tricks and no need for hess cycle only enthalpy change is required in this question: Ca(OH)2 + CO2 arrow CaCO3 + H2O ..... as the equation is hifted we change the sign of -82 to +82...and enthalpy change is products - reactants always remember that and

standard enthalpy change is reactants-products....then its a simple substraction did u get it? i hope it helps if not imma try again

got it...Thanks alot..Jazakillah. Its pretty easy..sometimes I get so dumb