工业合成尿素：$\text{HNCO}\left( \text{g} \right)\text{+N}{{\text{H}}_{\text{3}}}\left( \,\text{g} \right)\rightleftharpoons \text{CO}{{\left( \text{N}{{\text{H}}_{\text{2}}} \right)}_{\text{2}}}\left( \,\text{g} \right)\qquad\Delta\!\!\textit{ H}\lt 0$。恒容密闭容器加入等物质的量的$\text{HNCO}$和$\text{N}{{\text{H}}_{3}}$，在$\textit{T}_{1}$和$\textit{T}_{2}$反应达到平衡，$\text{lg}p\left( \text{N}\text{H}_{\text{3}}\right)$与$\text{lg}p\left[ \text{CO}\left( \text{N}\text{H}_{\text{2}}\right)_{\text{2}}\right]$关系如图所示$\rm ($压强单位均为$\text{kPa}\rm )$。反应$d\text{D}\left( \text{g}\right)+e\text{E}\left( \text{g}\right)\rightleftharpoons g\text{G}\left( \text{g}\right)$的标准平衡常数$\textit{K}^{\textit{o}}=\dfrac{\left[ \dfrac{\textit{p}\left( \text{G}\right)}{\textit{p}^{\textit{o}}}\right]^{\textit{g}}}{\left[ \dfrac{\textit{p}\left( \text{D}\right)}{\textit{p}^{\textit{o}}}\right]^{\textit{d}}\left[ \dfrac{\textit{p}\left( \text{E}\right)}{\textit{p}^{\textit{o}}}\right]^{\textit{e}}}，\textit{p}^{^{o}}=100\ \text{kPa}$。下列说法正确的是$(\qquad)$

$\\textit{T}_{\\text{1}}\\gt \\textit{T}_{\\text{2}}$

$\\dfrac{\\textit{K}_{\\text{p}}\\left( \\textit{T}_{\\text{1}}\\right)}{\\textit{K}_{\\text{p}}\\left( \\textit{T}_{\\text{2}}\\right)}=\\dfrac{\\textit{K}^{\\text{o}}\\left( \\textit{T}_{\\text{1}}\\right)}{\\textit{K}^{\\text{o}}\\left( \\textit{T}_{\\text{2}}\\right)}$

$\\textit{T}_{2}$下，平衡时体系的总压强为$250\\;\\rm \\text{kPa}$时，$\\textit{p}\\left( \\text{N}{{\\text{H}}_{\\text{3}}} \\right)\\approx 4.9\\;\\rm \\text{kPa}$

若改为恒压密闭容器，则平衡时$\\text{CO}{{\\left( \\text{N}{{\\text{H}}_{2}} \\right)}_{2}}$的体积分数减小

$\textit{p}\left( \text{N}\text{H}_{\text{3}}\right)$表示物质$\text{N}{{\text{H}}_{\text{3}}}$的分压强，$\text{lg}p\left( \text{N}\text{H}_{\text{3}}\right)$值越大，说明其分压越大，说明其体积分数或物质的量分数越大，据此分析解答。

$\rm A$．平衡时，相同的$\text{lg}p\left( \text{N}\text{H}_{\text{3}}\right)$时，温度${{\textit{T}}_{2}}$时的$\text{lg}p\left[ \text{CO}{{\left( \text{N}{{\text{H}}_{\text{2}}} \right)}_{\text{2}}} \right]$小于${{\textit{T}}_{1}}$时的$\text{lg}p\left[ \text{CO}{{\left( \text{N}{{\text{H}}_{\text{2}}} \right)}_{\text{2}}} \right]$  ，说明温度${{\textit{T}}_{1}}$到${{\textit{T}}_{2}\;\rm ^\circ\rm C}$时，平衡逆向移动，该反应是放热反应，则$\textit{T}_{\text{1}}\lt \textit{T}_{\text{2}}$，故$\rm A$错误；

$\rm B$．$\textit{K}^{\text{o}}=\dfrac{\left[ \dfrac{\textit{p}\left( \text{G}\right)}{\textit{p}^{\textit{o}}}\right]^{\textit{g}}}{\left[ \dfrac{\textit{p}\left( \text{D}\right)}{\textit{p}^{\textit{o}}}\right]^{\textit{d}}\left[ \dfrac{\textit{p}\left( \text{E}\right)}{\textit{p}^{\textit{o}}}\right]^{^{\textit{e}}}}=\dfrac{\textit{p}\left( \text{G}\right)^{\textit{g}}}{\textit{p}\left( \text{D}\right)^{\textit{d}}\textit{p}\left( \text{E}\right)^{\textit{e}}}\times\left[ \textit{p}^{\textit{o}}\right]^{\textit{d}+\textit{e}-\textit{g}}=\textit{K}_\rm{p}\left[ \textit{p}^{\textit{o}}\right]^{\textit{d}+\textit{e}-\textit{g}}$，则${{\textit{K}}^{\textit{o}}}\left( {{\textit{T}}_{\text{1}}} \right)\rm ={{\textit{K}}_{\text{p}}}\left( {{\textit{T}}_{\text{1}}} \right){{\left[ {{\textit{p}}^{\textit{o}}} \right]}^{\textit{d}+\textit{e}-\textit{g}}}$，${{\textit{K}}^{\text{o}}}\left( {{\textit{T}}_{\text{2}}} \right)\rm ={{\textit{K}}_{\text{p}}}\left( {{\textit{T}}_{\text{2}}} \right){{\left[ {{\textit{p}}^{\textit{o}}} \right]}^{\textit{d}+\textit{e}-\textit{g}}}$，故$\dfrac{\textit{K}_{\text{p}}\left( \textit{T}_{\text{1}}\right)}{\textit{K}_{\text{p}}\left( \textit{T}_{\text{2}}\right)}=\dfrac{\textit{K}^{\textit{o}}\left( \textit{T}_{\text{1}}\right)}{\textit{K}^{\textit{o}}\left( \textit{T}_{\text{2}}\right)}$，故$\rm B$正确；

$\rm C$．$\textit{T}_{2}$时，$\text{lg}p\left( \text{N}{{\text{H}}_{\text{3}}} \right)\rm =1$，$\text{lg}p\left[ \text{CO}{{\left( \text{N}{{\text{H}}_{\text{2}}} \right)}_{\text{2}}} \right]\rm =3$，则$\textit{p}\left( \text{N}{{\text{H}}_{\text{3}}} \right)\rm =10$，$\textit{p}\left( \text{HNCO}\right)=10$，$\textit{p}\left( \text{CO}{{\left( \text{N}{{\text{H}}_{\text{2}}} \right)}_{\text{2}}} \right)\rm =1000$，则$\textit{K}_\rm{p}\left( \textit{T}_{2}\right)=\dfrac{\textit{p}\left( \text{CO}\left( \text{N}\text{H}_{\text{2}}\right)_{\text{2}}\right)}{\textit{p}\left( \text{N}\text{H}_{\text{3}}\right)\textit{p}\left( \text{HNCO}\right)}=\dfrac{1000}{100}=10$，若平衡时，$\textit{p}\left( \text{N}{{\text{H}}_{\text{3}}} \right)\approx 4.9\;\rm \text{kPa}$，则$\textit{K}_\rm{p}\left(\textit{T}_{2}\right)=\dfrac{\textit{p}\left( \text{CO}\left( \text{N}\text{H}_{\text{2}}\right)_{\text{2}}\right)}{\textit{p}\left( \text{N}\text{H}_{\text{3}}\right)\textit{p}\left( \text{HNCO}\right)}=\dfrac{\textit{p}\left( \text{CO}\left( \text{N}\text{H}_{\text{2}}\right)_{\text{2}}\right)}{4.9\times4.9}=10$，则$\textit{p}\left( \text{CO}{{\left( \text{N}{{\text{H}}_{\text{2}}} \right)}_{\text{2}}} \right)\rm =240.1$ $\text{kPa}$，${{\textit{P}}_{总}}=\textit{p}\left( \text{CO}{{\left( \text{N}{{\text{H}}_{\text{2}}} \right)}_{\text{2}}} \right)+\textit{p}\left( \text{N}{{\text{H}}_{\text{3}}} \right)+\textit{p}\left( \text{HNCO} \right)=240.1+4.9+4.9=249.9\approx 250\ \text{kPa}$ ，则平衡时体系的总压强为$250\;\rm \text{kPa}$，故$\rm C$正确；

$\rm D$．此反应$\text{HNCO}\left( \text{g} \right)\text{+N}{{\text{H}}_{\text{3}}}\left( \ \text{g} \right)\rightleftharpoons \text{CO}{{\left( \text{N}{{\text{H}}_{\text{2}}} \right)}_{\text{2}}}\left( \ \text{g} \right)\ $是总物质的量减小的反应，若改为恒压密闭容器，相对于恒容密闭容器来说，相当于增大压强的反应，增大压强，平衡会正向移动，则若改为恒压密闭容器，平衡时$\text{CO}{{\left( \text{N}{{\text{H}}_{2}} \right)}_{2}}$的体积分数会增大，故$\rm D$错误；

故选：$\rm BC$