Question

the standard heats of formation at 298 k for ccl4,h20,co2 nad hcl are -25.5, -57.8, -94.1 and -22.1 k cal per mole respectively. then what is the at 298 k for the reaction,

Answer 1

Just take the difference between the enthalpy of formation of products and enthalpy of formation of reactants.

Take stoichiometric coefficients into consideration too.

$/Delta _r H = -94.1 - 4/times 22.1 -(-25.5 - 2\times 57.8)$

$/Delta _r H = 41.4$

Answer 2

Answer: The enthalpy  for the given reaction is -54.4 kCal/mol.

Explanation: Enthalpy change of the reaction is defined as the amount of heat released or absorbed in a given chemical reaction.

$\Delta H_{rxn}=\Delta H_f_{(products)}-\Delta H_f_{(reactants)}$

We are given a chemical reaction. The reaction follows:

$CCl_4+2H_2O\rightarrow 4HCl+CO_2$

Given:

$H_f_{(CCl_4)}=-25.5kCal/mol$

$H_f_{(H_2O)}=-57.8kCal/mol$

$H_f_{(CO_2)}=-94.1kCal/mol$

$H_f_{(HCl)}=-22.1kCal/mol$

Enthalpy change for the reaction of he given chemical reaction is given by:

$\Delta H_{rxn}=H_f_{(H_2O)}+H_f_{(CO_2)}-(H_f_{(CCl_4)}+H_f_{(H_2O)})$

Putting the values in above equation, we get

$\Delta H_{rxn}=[(94\times -22.1)+(-94.1)]-[(2\times -57.8)+(-25.5)]kCal/mol$

$\Delta H_{rxn}=-54.4kCal/mol$