Question

In a generic chemical reaction involving reactants A and B and products C and D, aA+bB→cC+dD, the standard enthalpy ΔH∘rxn of the reaction is given by ΔH∘rxn=cΔH∘f(C)+dΔH∘f(D) −aΔH∘f(A)−bΔH∘f(B) Notice that the stoichiometric coefficients, a, b, c, d, are an important part of this equation. This formula is often generalized as follows, where the first sum on the right-hand side of the equation is a sum over the products and the second sum is over the reactants: ΔH∘rxn=∑productsnΔH∘f−∑reactantsmΔH∘f where m and n represent the appropriate stoichiometric coefficients for each substance. Part A What is ΔH∘rxn for the following chemical reaction? H2O(l)+CCl4(l)→COCl2(g)+2HCl(g) You can use the following table of standard heats of formation (ΔH∘f) to calculate the enthalpy of the given reaction. Element/ Compound Standard Heat of Formation (kJ/mol) Element/ Compound Standard Heat of Formation (kJ/mol) H(g) 218 N(g) 473 H2(g) 0 O2(g) 0 CCl4(l) −139.5 O(g) 249 H2O(l) −285.8 HCl(g) −92.30kJ C(g) 71 COCl2(g) −218.8kJ C(s) 0 HNO3(aq) −206.6 Express the standard enthalpy of reaction to three significant figures and include the appropriate units.

Answer 1

given reaction is

H2O + CCl4 ---------> COCl2 + 2HCl

standard heat of formations of reactants DH H2O = -285.8 KJ/mol , DH CCl4 = -139.5 KJ/mol

standard heat of formations of products DH COCl2 = -218.8 KJ/mol , DH HCl = -92.3 KJ/mol

as per formula DH reaction = (sum of products standard heat of formation*number of moles presenting products)-(sum of reactants standard heat of formation*number of moles presenting reactants)= (DH COCl2*1 + DH HCl*2) - (DH H2O *1+ DH CCl4*1) = (-218.8 kj/mol*1mol-92.3 kj/mol*2)-(-285.8kj/mol-139.5kj/mol)*1mol

= 21.9 KJ

standard enthalpy of reaction = 21.9 KJ