Question

Calculate $K_p$ for the following reaction at ${25}^{\circ }C.$
$2N_{2}O\left(g\right)\rightleftharpoons 2N_2\left(g\right)+O_2\left(g\right)$
$\Delta G_f^0\left(N_{2}O\right)=104.2kJ/mol,\Delta G_f^0\left(N_2\right)=0and\Delta G_f^0\left(O_2\right)=0$

• A. $6.3\times {10}^{30}$
  
• B. $3.3\times {10}^{36}$
  
• C. $3.3\times {10}^{-36}$
  
• D. $6.3\times {10}^{-30}$
  

Answer 1

The correct option is B $3.3\times {10}^{36}$


We know for the reaction,
$\Delta G^o=\Delta G_f^o\left(products\right)-\Delta G_f^o\left(reactants\right)$
$\Delta G^0=[2\times \Delta G_f^0(N_2)+\Delta G_f^0(O_2\left)\right]-2\times \Delta G_f^0\left(N_{2}O\right)$
putting values,
$\Delta G^0=[2\times 0+0]-2\times 104200$
$\Delta G^0=-2.084\times {10}^{5}J/mol.$

This is a gas- phase reaction, so $\Delta G^0$ is related to $K_p$ by,
$\Delta G^0=-2.303RTlog{K}_p$
putting the values,
$-2.084\times {10}^5=-2.303\times 8.314\times 298log{K}_p$
Solving this,
$K_p=3.3\times {10}^{36}$