Question

6.80 g of $N{H}_3$; is passed over heated CuO. The standard heat enthalpies of $N{H}_3$, $CuO\left(s\right)$ and $H_{2}O\left(l\right)$ are -46.0, -155.0 and $-2850kJmo{l}^{-1}$ respectively and the change is.
$N{H}_3+\left(3/2\right)CuO⟶\frac{1}{2}{N}_2\left(g\right)+\left(3/2\right)H_{2}O\left(l\right)+\left(3/2\right)Cu\left(s\right)$.
If the enthalpy change is

• A. -50
• B. -59.6
• C. 60
• D. 30

Answer 1

The correct option is B -59.6

$N{H}_3+\left(3/2\right)CuO⟶\frac{1}{2}{N}_2\left(g\right)+\left(3/2\right)H2O\left(l\right)+\left(3/2\right)Cu\left(s\right);\Delta H=?$
$\Delta H^{\circ }=\Sigma H_{Products}^{\circ }-\Sigma H_{Reactants}^{\circ }$
$=[\frac{1}{2}\times H_{N_2}^{\circ }+(3/2)\times H_{H_{2}O\left(l\right)}^{\circ }+(3/2\left)H_{Cu}^{\circ }\right]-\left[\right(3/2)H_{CuO}^{\circ }+H_{N{H}_3}^{\circ }]$
$=[0+(3/2)\times (-285)+(3/2)\times 0]-\left[\right(3/2)\times (-155)+(-46\left)\right]$
$=-149kJ$
$[\because H^{\circ }forpureelement=0and{H}_{compound}^{\circ }=H_{formation}^{\circ }]$
$\because 17gN{H}_{3}bringsinchangein\Delta H=149kJ$
$\because 6.8gN{H}_{3}bringsinchangein\Delta H=(149\times 6.8)/17=59.6kJ$
$\therefore \Delta Hfor6.89gN{H}_3=-59.6kJ$