Question

the standard heat of formation of CH4, H2O and CH2OH are -76, -242 and -266kj/mol respectively. the enthalpy change for the following reaction is; CH2OH(L)+H2(g)-CH4(g)+H2O(L)

Answer 1

Answer: The enthalpy change for the reaction is -52kJ/mol.

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 Hess’s law, the chemical equation can be treated as algebraic expressions 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 standard heat of formation is the enthalpy change associated with the formation of one mole of a compound from its elements in their standard states.

$C(s)+2H_2(g)\rightarrow CH_4(g)$  $\Delta H_1=-76kJ/mol$ (1)

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$  $\Delta H_2=-242kJ/mol$ (2)

$C(s)+2H_2(g)+\frac{1}{2}O_2(g)\rightarrow CH_3OH(l)$  $\Delta H_3=-266kJ/mol$ (3)

Adding 1 and 2 and subtracting 3 , the net chemical equation:

$CH_3OH(l)+H_2(g)\rightarrow CH_4(g)+H_2O(l)$  $\Delta H=?$ (4)

$\Delta H_4=\Delta H_1+\Delta H_2-\Delta H_3=(-76)+(-242)-(-266)=-52kJ/mol$

The enthalpy change for the reaction is -52kJ/mol.