• # question_answer The reaction between ${{N}_{2}}$ and ${{H}_{2}}$ to form ammonia has ${{K}_{c}}=6\times {{10}^{-2}}$ at the temperature 500°C. The numerical value of ${{K}_{p}}$ for this reaction is  A) $1.5\times {{10}^{-5}}$ B) $1.5\times {{10}^{5}}$ C) $1.5\times {{10}^{-6}}$ D) $1.5\times {{10}^{6}}$

 [a] ${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}$; $\Delta n=2-4=-2$ ${{K}_{p}}=6\times {{10}^{-2}}\times {{(0.0812\times 773)}^{-2}}$ ${{K}_{p}}=\frac{6\times {{10}^{-2}}}{{{(0.0812\times 773)}^{2}}}=1.5\times {{10}^{-5}}$.