Given the following thermochemical equations: | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The enthalpy of formation of $H_2SO_4$ corresponds to the reaction: $H_2 + S + 2O_2 \rightarrow H_2SO_4$.To obtain this,we add the given equations

Vedclass · Accepted Answer

$-814.4$

Vedclass · Answer

(A) The enthalpy of formation of $H_2SO_4$ corresponds to the reaction: $H_2 + S + 2O_2 \rightarrow H_2SO_4$.To obtain this,we add the given equations:$(1) S + O_2 \rightarrow SO_2 ; \Delta H_1 = -298.2 \ kJ$$(2) SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 ; \Delta H_2 = -98.7 \ kJ$$(3) SO_3 + H_2O \rightarrow H_2SO_4 ; \Delta H_3 = -130.2 \ kJ$$(4) H_2 + \frac{1}{2} O_2 \rightarrow H_2O ; \Delta H_4 = -287.3 \ kJ$Summing these equations:$(S + O_2) + (SO_2 + \frac{1}{2} O_2) + (SO_3 + H_2O) + (H_2 + \frac{1}{2} O_2) \rightarrow SO_2 + SO_3 + H_2SO_4 + H_2O$Canceling common terms on both sides:$S + H_2 + (1 + \frac{1}{2} + \frac{1}{2}) O_2 \rightarrow H_2SO_4$$S + H_2 + 2O_2 \rightarrow H_2SO_4$The total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4$$\Delta H = (-298.2) + (-98.7) + (-130.2) + (-287.3) = -814.4 \ kJ$.

Bond	Bond energy $(kJ \ mol^{-1})$
$C-H$	$413$
$C-C$	$348$
$C=C$	$610$
$H-H$	$436$

Similar Questions

The conversion of oxygen to ozone represented by the equation $3O_2 \to 2O_3$ is an endothermic reaction. The enthalpy change $\Delta H$ accompanying the reaction:

What amount of energy (in $KJ$) is released in the combustion of $5.8\, g$ of $C_4H_{10}\,(g)$?
$2 C_4H_{10}(g) + 13 O_2(g) \to 8 CO_2(g) + 10 H_2O(l)$; $\Delta H^o = -5756\, KJ$

Given:
$I$. $2Fe_{(s)} + \frac{3}{2} O_{2(g)} \to Fe_2O_{3(s)}$; $\Delta H^{\Theta} = -193.4 \ kJ$
$II$. $Mg_{(s)} + \frac{1}{2} O_{2(g)} \to MgO_{(s)}$; $\Delta H^{\Theta} = -140.2 \ kJ$
What is $\Delta H^{\Theta}$ of the reaction?
$3Mg_{(s)} + Fe_2O_{3(s)} \to 3MgO_{(s)} + 2Fe_{(s)}$
....... $kJ$

The heat of neutralization of a strong acid with a strong base is constant and is equal to:

With the help of the following data,find out the change in heat content for the reaction in $kJ$:
$C_2H_{4(g)} + H_{2(g)} \to C_2H_{6(g)}$

Bond Bond energy $(kJ \ mol^{-1})$

$C-H$ $413$

$C-C$ $348$

$C=C$ $610$

$H-H$ $436$

Vedclass Test Series

Exam Paper Generator

Online Exam Module