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Question: The \(\Delta H_f^0\) for \(C{O_2}(g)\), \(CO(g)\) and \({H_2}O(g)\) are \( - 393.5, - 110.5\) and \(...

The ΔHf0\Delta H_f^0 for CO2(g)C{O_2}(g), CO(g)CO(g) and H2O(g){H_2}O(g) are −393.5,−110.5 - 393.5, - 110.5 and −241.8kJmol−1 - 241.8kJmo{l^{ - 1}} respectively. The standard enthalpy change (in kJkJ) for the reaction –
CO2(g)+H2(g)→CO(g)+H2O(g)C{O_2}(g) + {H_2}(g) \to CO(g) + {H_2}O(g) is
A.324.1
B.41.2
C.-262.5
D.-41.2

Explanation

Solution

Here, in this question we have to find the standard enthalpy change. Firstly standard enthalpy of a given reaction or product is the measure of energy released or consumed when one mole of the substance is created under standard conditions. The standard conditions are 298.15K temperature and 1atm pressure. First we shall find out the enthalpy of the product and then the enthalpy of the reactant. Its difference will give us the required solution.

Complete answer:
Given data contains,
ΔHf0\Delta H_f^0 of CO2(g)C{O_2}(g) = −393.5kJ/mol - 393.5kJ/mol
ΔHf0\Delta H_f^0 of CO(g)CO(g) = −110.5kJ/mol - 110.5kJ/mol
ΔHf0\Delta H_f^0 of H2O(g){H_2}O(g) = −241.8kJ/mol - 241.8kJ/mol
The chemical reaction takes place in the following manner –
CO2(g)+H2(g)→CO(g)+H2O(g)C{O_2}(g) + {H_2}(g) \to CO(g) + {H_2}O(g)
Reactants are – CO2(g)C{O_2}(g) and H2(g){H_2}(g)
Products are – CO(g)CO(g) and H2O(g){H_2}O(g)
First, we will calculate the enthalpy of reactants. The equation is as follows:
ΔH(reactants)∘=ΔHf0(CO2)+ΔHf0(H2)\Delta H_{\left( {{\text{reactants}}} \right)}^\circ = \Delta H_f^0(C{O_2}) + \Delta H_f^0({H_2})
Since ΔHf0(H2)=0\Delta H_f^0({H_2}) = 0, under standard conditions.
ΔH(reactants)0\Delta H_{({\text{reactants}})}^0= −393.5+(0) - 393.5 + (0)
= −393.5kJ/mol - 393.5kJ/mol
Now, we will calculate the enthalpy of products. The equation is as follows:
ΔH(products)∘=ΔHf0(CO)+ΔHf0(H2O)\Delta H_{\left( {{\text{products}}} \right)}^\circ = \Delta H_f^0(CO) + \Delta H_f^0({H_2}O)
Now we substitute the known values we get,
= −110.5+(−241.8) - 110.5 + ( - 241.8)
= −110.5−241.8 - 110.5 - 241.8
On simplification we get,
= −352.3kJ/mol - 352.3kJ/mol
To find the standard enthalpy change, we will evaluate using the following equation:
ΔH(reaction)0=ΔH(product)o−ΔH(reactant)0\Delta H_{(reaction)}^0 = \Delta H_{{\text{(product)}}}^o - \Delta H_{({\text{reactant}})}^0
On substituting the known values we get,
⇒ΔH(reaction)0=(−393.5)−(−352.3)\Rightarrow \Delta H_{{\text{(reaction)}}}^0 = ( - 393.5) - ( - 352.3)
⇒ΔH(reaction)0=(−393.5+352.3)\Rightarrow \Delta H_{{\text{(reaction)}}}^0 = ( - 393.5 + 352.3)
On simplification we get,
ΔH(reaction)0=−41.2kJ/mol\Delta H_{(reaction)}^0 = - 41.2kJ/mol
Hence, the correct answer is option (D).

Note:
We need to know that the chemical reaction takes place in either endothermic, exothermic or neutral. In endothermic reactions, the enthalpy of reactant is greater than the enthalpy of products. This question particularly is a type of exothermic reaction where the enthalpy also called energy of the reactants are lesser than the energy of the products. We obtain the enthalpy change by the standard formula, where we determine the change as a difference between the enthalpy of product and reactants.