Question

The standard enthalpy change for the following reaction is 232 kJ at 298 K. 2 H2CO(g) 2 C(s,graphite) + 2 H2(g) + O2(g) ΔH° = 232 kJ What is the standard enthalpy change for this reaction at 298 K? C(s,graphite) + H2(g) + 1/2 O2(g) H2CO(g)

Answer 1

Answer:  The standard enthalpy change for this reaction is -116 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

$2H_2CO(g)\rightarrow 2C(s, graphite)+2H_2(g)+O_2(g)$

$\Delta H^0=232KJ$

Now we have to determine the value of $\Delta H$ for the following reaction i.e,

$)C(s, graphite)+H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2CO(g)$ $\Delta H^0'=?$

According to the Hess’s law, if we reverse the reaction then the $\Delta H$ will change its sign and if we half the reaction, then the

So, the value $\Delta H^0'$ for the reaction will be:

$\Delta H^0'=\frac{(-232kJ/mole)}{2}=-116kJ$

Hence, the standard enthalpy change for this reaction is -116 kJ