Question

For each of the following reactions, the enthalpy change given is that measured when the numbers of moles of reactants and products taking part in the reaction are as given by their coefficients in the equation. Calculate the enthalpy change whengram of the underlined substance is consumed or produced.

1. $\mathbf{\text{Ca}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\underset{\mathbf{¯}}{{\mathbf{\text{Br}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{\ell }\mathbf{\right)}}\mathbf{\to }{\mathbf{\text{CaBr}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{\Delta }\mathbf{H}\mathbf{=}\mathbf{-}\mathbf{683}\mathbf{}\mathbf{\text{kJ}}$
2. $\mathbf{6}{\mathbf{\text{Fe}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{4}\underset{\mathbf{¯}}{{\mathbf{\text{Fe}}}_{\mathbf{3}}{\mathbf{\text{O}}}_{\mathbf{4}}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\Delta }\mathbf{H}\mathbf{=}\mathbf{\text{472}}\mathbf{ }\mathbf{\text{KJ}}$
3. $\mathbf{2}\underset{\mathbf{¯}}{{\mathbf{\text{NaHSO}}}_{\mathbf{4}}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{\text{NaOH}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{\text{SO}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{,}\mathbf{\Delta }\mathbf{H}\mathbf{=}\mathbf{+}\mathbf{806}\mathbf{}\mathbf{\text{kJ}}$

Short answer

(a) The enthalpy changes for the production of$1.00\text{g}{\text{Br}}_2$is$-4.27\text{kJ}$.

(b) The enthalpy change for the production of$1.00\text{g}{\text{Fe}}_3{\text{O}}_4$is$+0.509\text{kJ}$.

(c) The enthalpy change for the production of $1.00\text{g}{\text{NaHSO}}_4$ is $+3.36\text{kJ}$.

Step-by-step solution

Concept of thermochemistry

The study of heat energy related to chemical reactions and/or phase changes such as melting and boiling is known as thermochemistry.

(a) Calculation of enthalpy for 1 g  of  Br2 

$Ca\left(s\right)+\underset{¯}{B{r}_2\left(l\right)}\to CaB{r}_2\left(s\right)$

The enthalpy change for the above reaction , $\Delta H=-683\text{kJ}$.

Thus is released for the consumption of 1 mole of ${\text{Br}}_{\text{2}}$.

$\text{Molar} \text{mass} \text{of} {\text{Br}}_2\text{,M}\left({\text{Br}}_{\text{2}}\right)=159.80\frac{\text{g}}{\text{mol}}$
The enthalpy change for $159.80 {\text{g of Br}}_2\text{= - 683 KJ}$

The enthalpy change for $1 \text{g of} {\text{Br}}_2\text{=}\frac{\text{- 683} \text{KJ}}{\text{159.80g}}$

= $-4.27 \text{KJ}$

(b) Calculation enthalpy for 1g of Fe3O4  

$6F{e}_2{O}_3\left(s\right)\to 4\underset{¯}{F{e}_3{O}_4}\left(s\right)+O_2\left(g\right)$

It is given that, $\Delta H=+472\text{kJ}$. $472\text{kJ}$is absorbed for the production of 4 moles of ${\text{Fe}}_3{\text{O}}_4$.

$\text{Molar} \text{mass} \text{of} {\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}{\text{,M(Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{) = 231.6}\frac{\text{g}}{\text{mol}}$

$$\begin{gathered} {\text{Mass of 4 moles of Fe}}_3{\text{O}}_4\text{= 4} \text{mol}\times \text{231.6} \text{g/mol} \\ \text{= 926.4} \text{g} \end{gathered}$$

The enthalpy change for $926.4{\text{g of Fe}}_3{\text{O}}_4\text{= + 472 KJ}$

The enthalpy change for $1 {\text{g of Fe}}_3{\text{O}}_4\text{=}\frac{+472 \text{KJ}}{\text{926.4g}}$

= $0.509 \text{KJ}$

(c) Calculation of enthalpy for  1 g  of  NaHSO4

$2\underset{¯}{NaHS{O}_4}\left(s\right)\to 2NaOH\left(s\right)+2S{O}_2\left(g\right)+O_2\left(g\right)$

It is given that, $\Delta H=+806\text{kJ}$.

$806 \text{KJ}$ is absorbed for the consumption of 2 moles of ${\text{NaHSO}}_4$.

$\text{Molar} \text{mass} \text{of} {\text{NaHSO}}_{\text{4}}{\text{,M(NaHSO}}_{\text{4}}\text{) = 120.1}\frac{\text{g}}{\text{mol}}$

$$\begin{gathered} {\text{Mass of 2 moles of NaHSO}}_4\text{= 2} \text{mol}\times \text{120.1} \text{g/mol} \\ \text{= 240.2} \text{g} \end{gathered}$$

The enthalpy change for $240.2{\text{g of NaHSO}}_4\text{= + 806 KJ}$

The enthalpy change for $1 {\text{g of NaHSO}}_4\text{=}\frac{+806 \text{KJ}}{\text{240.2g}}$

= $3.36 \text{KJ}$