Question

Use the information given here, data from Appendix D, and equation (7.22) to calculate the standard enthalpy of formation per mole of ZnS(s). 2 ZnS(s) + 302(g) —2 ZnO(s) + 2 SO2(g) A Hº = -878.2 kJ mol-1 (Reference to Equation 7.22):

Answer 1

You have not provide appendix D and equation 7.22

So, I have solved the problem myself.

Am. we have, 2 Inscs) + 30209) 7 2 Znocs, + 250209). AH = -878.2 KJ, DHp =-348 OH = -296.1 Imol . WIlmor u mal Now, AHAM = LAHA coroducts - EoHFcreacteint -02 SAHEB Oudert-2(AHzno) + 2 CARP 50? = 2 (-348 kq) + 2(-296.1 ha) moll - 696 K + (-592.2 KB mol , mol =-1288.2 KB mol Now, DHE Ens - (9.) und DHA 02:0. Edhe soeuctens) = 2 ( orię zns ) + 3( shef oz) = 2 (AHans) + 3 (0) = 2(ongens)

According to eq??). mol mol 3 -1288.2 k] + 878.2 ] mol mol -878,2 kJ = -1288.2 KJ - 2 (otec Ins) :. 2 (DHørens) = -1288.2 K2, + 878.2 had • 2 compens) = -410 is stig ens = -410 kJ/mol is oll'ans - 205 kJ 2 17 Ens - 205 mol

Standard enthalpy of formation per mole of ZnS is -205 KJ/mol.

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