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Thermodynamics

# Given that, $NH_3 + 3Cl_2(g) \rightarrow NCl_3(g) + 3HCl(g);\: \: \: \: \Delta H_1 = -x_1$  $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g);\: \: \: \: \: \: \: \: \Delta H_2 = -x_2$  $H_2(g) + Cl_2(g) \rightarrow 2HCl(g);\: \: \: \: \: \: \: \: \Delta H_3 = x_3$  The heat of formation of $NCl_3(g)$ from the above data is

$\begin {array} {1 1} (A)\;–x_1 + x_2/2 – x_3 & \quad (B)\;x_1 + x_2/2 – x_3 \\ (C)\;x_1 - x_2/2 – x_3 & \quad (D)\;–x_1 - x_2/2 – x_3 \end {array}$

Can you answer this question?

$\large\frac{1}{2}$$N_2 + Cl_2 \rightarrow NCl_3; \Delta H$
$\Delta H = \Delta H_1+\large\frac{1}{2} \Delta H_2 - \Delta H_3 = -x_1 – x_2/2 – 3x_3/2$
Ans : (D)
answered Mar 17, 2014