Question

a chemist measures the enthalpy change \\( \delta h \\) during the following reaction:

\\( 2 \mathrm { hno } _ { 3 } ( l ) + \mathrm { mg } ( s ) \
ightarrow \mathrm { mg } \left( \mathrm { no } _ { 3 } \
ight) _ { 2 } ( s ) + \mathrm { h } _ { 2 } ( g ) \quad \delta h = - 443. \mathrm { kj } \\)

use this information to complete the table below. round each of your answers to the nearest \\( \mathrm { kj } \\).

reaction	\\( \delta h \\)

| \\( \mathrm { mg } \left( \mathrm { no } _ { 3 } \
ight) _ { 2 } ( s ) + \mathrm { h } _ { 2 } ( g ) \
ightarrow 2 \mathrm { hno } _ { 3 } ( l ) + \mathrm { mg } ( s ) \\) | \\( \square \mathrm { kj } \\) |
| \\( 6 \mathrm { hno } _ { 3 } ( l ) + 3 \mathrm { mg } ( s ) \
ightarrow 3 \mathrm { mg } \left( \mathrm { no } _ { 3 } \
ight) _ { 2 } ( s ) + 3 \mathrm { h } _ { 2 } ( g ) \\) | \\( \square \mathrm { kj } \\) |
| \\( \frac { 1 } { 2 } \mathrm { mg } \left( \mathrm { no } _ { 3 } \
ight) _ { 2 } ( s ) + \frac { 1 } { 2 } \mathrm { h } _ { 2 } ( g ) \
ightarrow \mathrm { hno } _ { 3 } ( l ) + \frac { 1 } { 2 } \mathrm { mg } ( s ) \\) | \\( \square \mathrm { kj } \\) |

Explanation:

Step1: Reverse the reaction

When a reaction is reversed, the sign of $\Delta H$ changes.
For the reaction \(Mg(NO_3)_2(s)+H_2(g)\to2HNO_3(l) + Mg(s)\), it is the reverse of \(2HNO_3(l)+Mg(s)\to Mg(NO_3)_2(s)+H_2(g)\). So \(\Delta H = 443\space kJ\)

Step2: Multiply the reaction by a factor

When a reaction is multiplied by a factor \(n\), \(\Delta H\) is also multiplied by \(n\).
For the reaction \(6HNO_3(l)+3Mg(s)\to3Mg(NO_3)_2(s)+3H_2(g)\), the original reaction \(2HNO_3(l)+Mg(s)\to Mg(NO_3)_2(s)+H_2(g)\) is multiplied by \(n = 3\). So \(\Delta H=3\times(- 443\space kJ)=-1329\space kJ\)

Step3: Multiply the reaction by a fraction

When a reaction is multiplied by a factor \(n=\frac{1}{2}\), \(\Delta H\) is also multiplied by \(n\).
For the reaction \(\frac{1}{2}Mg(NO_3)_2(s)+\frac{1}{2}H_2(g)\to HNO_3(l)+\frac{1}{2}Mg(s)\), the reverse of the original reaction \(2HNO_3(l)+Mg(s)\to Mg(NO_3)_2(s)+H_2(g)\) is multiplied by \(n=\frac{1}{2}\). So \(\Delta H=\frac{1}{2}\times443\space kJ = 221.5\approx222\space kJ\)

Answer:

reaction	\(\Delta H\)
\(6HNO_3(l)+3Mg(s)\to3Mg(NO_3)_2(s)+3H_2(g)\)	\(-1329\space kJ\)
\(\frac{1}{2}Mg(NO_3)_2(s)+\frac{1}{2}H_2(g)\to HNO_3(l)+\frac{1}{2}Mg(s)\)	\(222\space kJ\)