Question

Evaluate Yourself - 1
Calculate ∆H° for the reaction
CO2(g)+ H2(g) → CO(g)+ H2O(g)
given that AH' for CO2 (g), CO (g) and
H2O(g) are - 393.5, - 111.31 and - 242
kJ mol-1 respectively​

Answer 1

Answer:

The standard formation enthalpy for the given reaction is $+\ 40.19$ $KJmol^{-1}$·

Explanation:

The balanced chemical equation  is given as,

   $CO_2_{(g)} + H_2_{(g)}\ --->\ CO_{(g)} \ +\ H_2O_{(g)}$

Given that,

The standard formation enthalpy, $\Delta_{f} H^{0}$ of $CO_2$  $=$  $-393.5\ KJ mol^{-1}$

The standard formation enthalpy,  $\Delta_{f} H^{0}$ of $CO$  $=$  $-111.31$ $KJmol^{-1}$

The  standard formation enthalpy, $\Delta_{f} H^{0}$ of $H_2O$  $=$  $-242$ $KJmol^{-1}$

We know that,

The formula to calculate the standard formation enthalpy follows,

$\Delta_{f}H^{0}$  $=$ $\Sigma \Delta_{f}H^{0}_{(prodicts)} \ -\ \Sigma \Delta_{f}H^{0}_{(reactants)}$

$\Delta_{f}H^{0}$   $=$ $\Delta_{f}H^{0}_{(CO)} \ + \Delta_{f}H^{0}_{(H_2O)} \ -\ [\ \Delta_{f}H^{0}_{(CO_2)\ +\ }$ $\Delta_{f}H^{0}_{(H_2)$ $]$

On substituting the values we get,

 ⇒  $\Delta_{f}H^{0}$   $=$ $-111.31\ +\ (-242)\ -\ [\ -393.5\ +\ 0\ ]$

 ⇒  $\Delta_{f}H^{0}$   $=$ $-353.31\ +\ 393.5$

 ⇒  $\Delta_{f}H^{0}$   $=$ $+\ 40.19$ $KJmol^{-1}$

Therefore,

The standard formation enthalpy for the given reaction is $+\ 40.19$ $KJmol^{-1}$·