Question

A scientist measures the standard enthalpy change for the following reaction to be -56.2 kJ :

Ca(OH)₂(aq) + 2 HCl(aq)CaCl₂(s) + 2 H₂O(l)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CaCl₂(s) is   kJ/mol.

Answer in kJ/mol

Answer 1

Answer! We know that the standard enthalpy change of a balanced edemical reaction is the sum of enthalpies of the products minus the enthalpies of the reactants. Here we must take into account the coefficients of each components, Mathematically it can be expressed as, Home In Woº (product) – Im Hell (reactants) Where, n and on are coefficients. The given reaction is as follows: Cacon), cad) + 2HO (ad) + Cadiz (s) + 2400 he know, Hi [cacol) Call7 = - 985'2 KJ/MOL s HeI Cad)] = -164-2 kj/mol SHE [HOW] = - 285.829 kJ/mok L ui DH X = - 56.2 kJ / mer so we can write, - 56.2 K3 /mol = golf [cally (s)] + 2x SH 1,001}} - Low [cacom)cad] 0V, - 56.2 *J/mol = { swi [callce)] + 2x(-285-829 xs/mon}} + 2x <Hffice ad)]} -{ – 985•2 kJ/mol + 2x(-10712 45/me)} 9, - 56.2 kJ/mol = {cx> [cack, cs] – 571.658 ks]moe} - 4 - 985.2 ks/mok : - 334.4 kJ) mer? or, - 56.2. 89/ med = q suº [caus« so] - 571.658 ks/mos} - (- 1319.6x3]mok) 9,-6H: [cally(s)] = + 56.2 x5 /mok - 571.658 kilmol + 1319.6 kJ/mos 0,- GHz [callz (s)] = 804.142 ks/mol .She[cacy (s)] = - 804.142 kJ/mol The standard enthalpy of formation of cach is - 804.142 kJ/mol. ( Answer