Enthalpies of formation of CCl_4( g),H_2O(g),CO_2( g) and HCl are -105,-242,-394 and -92 kJ mol^(-1) respectively. The magnitude of enthalpy of the…

Question

Enthalpies of formation of CCl_4( g),H_2O(g),CO_2( g) and HCl are -105,-242,-394 and -92 kJ mol^(-1) respectively. The magnitude of enthalpy of the reaction given below is kJmol^(-1). (nearest integer) CCl_4( g)+2H_2O(g)→CO_2( g)+4HCl(g)

MARKS Content Team · Accepted Answer

Correct answer: 173. Using Hess's law of constant heat summation: Enthalpy of reaction = Enthalpy of formation of products - Enthalpy of formation of reactants ∴ΔH_f°=ΔH_f°CO_2, g+4ΔH_f°(HCl)-ΔH_f°CCl_4-2ΔH_f°H_2O,r =[-394]+4[-92]+105-2×[242] =-394-368+105+484 =-173 kJ/mole

Enthalpies of formation of ${CCl}_{4} (g), H_{2} O (g), {CO}_{2} (g)$ and $HCl$ are $- 105, - 242, - 394$ and $- 92 kJ {mol}^{- 1}$ respectively. The magnitude of enthalpy of the reaction given below is ${kJmol}^{- 1}$ . (nearest integer) ${CCl}_{4} (g) + 2 H_{2} O (g) \to {CO}_{2} (g) + 4 HCl (g)$

Solution

Relevant Questions

Enthalpies of formation of CCl4( g),H2O(g),CO2( g) and HCl are -105,-242,-394 and -92 kJ mol-1 respectively. The magnitude of enthalpy of the reaction given below is kJmol-1. (nearest integer) CCl4( g)+2H2O(g)→CO2( g)+4HCl(g)

Solution

Relevant Questions

Enthalpies of formation of ${CCl}_{4} (g), H_{2} O (g), {CO}_{2} (g)$ and $HCl$ are $- 105, - 242, - 394$ and $- 92 kJ {mol}^{- 1}$ respectively. The magnitude of enthalpy of the reaction given below is ${kJmol}^{- 1}$ . (nearest integer) ${CCl}_{4} (g) + 2 H_{2} O (g) \to {CO}_{2} (g) + 4 HCl (g)$