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

Question

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