Question

Enthalpies of formation of $CCl _4( g ), H _2 O ( g ), CO _2( g )$ and $HCl ( g )$ are $-105,-242,-394$ and $-92\, kJ$ mol $^{-1}$ respectively The magnitude of enthalpy of the reaction given below is ___$kJ \, mol ^{-1}$ (nearest integer) $CCl _4( g )+2 H _2 O ( g ) \rightarrow CO _2( g )+4 HCl ( g )$

Hint:

To calculate enthalpy change for a reaction, use the formula \( \Delta H = \sum H_p - \sum H_R \), where \( H_p \) and \( H_R \) are the enthalpies of products and reactants, respectively.

Answer 1

Correct Answer: 173

To find the enthalpy change for the reaction: $CCl_4(g) + 2H_2O(g) \rightarrow CO_2(g) + 4HCl(g)$, we use the standard enthalpies of formation ($\Delta_fH^\circ$) given: $CCl_4(g) = -105\,kJ/mol$, $H_2O(g) = -242\,kJ/mol$, $CO_2(g) = -394\,kJ/mol$, $HCl(g) = -92\,kJ/mol$. The enthalpy change ($\Delta_rH^\circ$) for the reaction is calculated using the formula:

$\Delta_rH^\circ = \Sigma (\Delta_fH^\circ \text{ of products}) - \Sigma (\Delta_fH^\circ \text{ of reactants})$

First, calculate the total enthalpy of formation for the products:

$1 \times (-394) + 4 \times (-92) = -394 - 368 = -762\,kJ/mol$

Next, calculate the total enthalpy of formation for the reactants:

$1 \times (-105) + 2 \times (-242) = -105 - 484 = -589\,kJ/mol$

Now, substitute these values into the formula:

$\Delta_rH^\circ = -762 - (-589) = -762 + 589 = -173\,kJ/mol$

The magnitude of the enthalpy change is $173\,kJ/mol$. Thus, the calculated value ($173\,kJ/mol$) matches the given range of $173\,kJ/mol$, confirming the solution.