Print Cakulator -a Periodic Table of 9 pling Learning Given the following inform

Question

Print Cakulator -a Periodic Table of 9 pling Learning Given the following information AH -738.2 kJ AH-458.0 kJ 5.-289.0 J / K AS-189.0 J/K A+B 2D calculate for the following reaction at 298 K. A+B 2C Number kJ University Science Books The tollowing equation represents the decomposition of a generic diatomic element in its standand state Assume that the standard molar Gibbs energy of formation of X(g) is 4.42 J mol" at 2000 K and -57 58 kJ mot at 3000. K Determine the value of K (the thermodynamic equilbrium oant) at each Number K at 2000. K-O Number K at 3000. K- r is independent of temperature, detemine the value of from these daa

Explanation / Answer

Given

A+B->2D, DeltaH=738.2 Kj, deltas= 289J/K (1)

C ----->D, deltaH= 458KJ and deltas= -189 J/K (2)

When Eq.2 is reversed, the sign of deltaH and deltas also gets revesed

D--->C, deltas= -458 KJ and deltas= 189J/K

Reversing the second reaction and multiplying with 2 gives 2D----->2C, deltaH= -458*2=-916KJ and deltas= 2*189=378 j/K (2A).

Eq.1+Eq.2A gives A+B----->2C, deltaH= 738.2-916 =-177.8 KJ and deltas=289+378=667J/K

the gibbs free energy change (deltaG0) of desire reaction.

2. since deltaG= -RT lnK, K= -deltaG/RT, R = 8.314 J/mole.K

at T1= 2000K, given deltaGo=4.42 Kj/mole and lnK1= -4.42*1000/(2000*8.314)=-0.27 and K1= 0.77

at T2= 3000K, deltaG= -57.6 Kj/mole and lnK2= 57.6*1000/(3000*8.314)=2.31 and K2= 10.1

since deltaH is independent of temperature

ln(K2/K1)= (deltaH/R)*(1/T1-1/T2), R = 8.314 J/mole

ln(10.1/0.77)= (deltaH/R)*(1/2000-1/3000)

deltaH= 128396.4 J/mole =128.4 Kj/mole the reaction is endothermic.

deltaGo= deltaH-T*deltas= -177.8 KJ- (298* 667J* 1KJ/1000J) = -376.6KJ

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