Question

Determine the enthalpy change for the reaction 2C(s) + 2H2O(g) → CH4(g) + CO2(g) using the following:

Answer 1

Answer:

The answer is actually -856.7 kJ

Explanation:

That's what it is on ed-genuity.

Answer 2

Answer : The enthalpy change for the reaction is, 97.7 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 main chemical reaction is,

$2C(s)+3H_2O(g)\rightarrow CH_4(g)+CO_2(g)$    $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)$     $\Delta H_1=131.3kJ$

(2) $CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)$    $\Delta H_2=41.2kJ$

(3) $CH_4(g)+H_2O(g)\rightarrow 2H_2(g)+CO(g)$    $\Delta H_3=206.1kJ$

Now we are multiplying reaction 1 by 2 and reversing reaction 3 and then adding all the equations, we get :

(1) $2C(s)+2H_2O(g)\rightarrow 2CO(g)+2H_2(g)$     $\Delta H_1=2\times 131.3kJ=262.6kJ$

(2) $CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)$    $\Delta H_2=41.2kJ$

(3) $2H_2(g)+CO(g)\rightarrow CH_4(g)+H_2O(g)$    $\Delta H_3=-206.1kJ$

The expression for enthalpy of main reaction will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(262.6)+(41.2)+(-206.1)$

$\Delta H=97.7kJ$

Therefore, the enthalpy change for the reaction is, 97.7 kJ