When S in the form of ${{\text{S}}_{8}}$ is heated at 900K, the initial pressure of 1 atmosphere falls by 29% at equilibrium. This is because of conversion of some ${{\text{S}}_{8}}\left( \text{g} \right)$ to ${{\text{S}}_{2}}\left( \text{g} \right)$. The ${{\text{K}}_{\text{p}}}$ for the reaction is x. Then 100x is _________.

Hint: The question deals in with the partial pressure terms ${{\text{K}}_{\text{p}}}$ is the equilibrium constant calculated from the partial pressures of a chemical reaction. It is a unitless number. Balance the reaction and use the value of degree of dissociation to find partial pressures and then ${{\text{K}}_{\text{p}}}$ .

Complete step by step answer:
Let us discuss this numerical step by step:
Step (1)- We need to balance the reaction of decomposition of ${{\text{S}}_{8}}\left( \text{g} \right)$. The atoms in the left hand side are 8 and in the right hand side are 2. To maintain the balance of atoms, the right side should have 8 atoms of sulphur. So, multiply ${{\text{S}}_{2}}\left( \text{g} \right)$ with 4 to get required reaction; ${{\text{S}}_{8}}\longrightarrow 4{{\text{S}}_{2}}$.

Step (2)- Write the formula: The formula ${{\text{K}}_{\text{p}}}$ for is $\dfrac{\text{product}{{\text{s}}_{(\text{p)}}}}{\text{reactant}{{\text{s}}_{(\text{p)}}}}$. To find ${{\text{K}}_{\text{p}}}$ we need to find $\text{product}{{\text{s}}_{(\text{p)}}}$ and $\text{reactant}{{\text{s}}_{\text{(p)}}}$.

Step (3)- Find partial pressures and degree of dissociation: For that, we need to take initial and equilibrium pressures into account.
At t=0, the pressure is 1 atmosphere which is given.
At equilibrium, $\alpha $ part is dissociated, then, the amount of ${{\text{S}}_{8}}\text{ left is }$ and ${{\text{S}}_{2}}\text{ formed will be}$.

The pressures will be $\begin{align}
  & {{\text{S}}_{8}}={{\text{P}}_{0}}\left( 1-\alpha \right) \\
 & {{\text{S}}_{2}}=4{{\text{P}}_{0}}\alpha \\
\end{align}$ , where${{\text{P}}_{0}}$is initial pressure and$\alpha $ is the degree of dissociation.
The value of degree of dissociation is $\alpha $ which is given as (falls by 29%). So, $\alpha =\dfrac{29}{100}$, which is equal to $\alpha =0.29$.
Hence, the ${{\text{K}}_{\text{p}}}$ is $\dfrac{{{\left( 4{{\text{P}}_{0}}\alpha \right)}^{4}}}{{{\text{P}}_{0}}\left( 1-\alpha \right)}$.

Step (4)- Put the values of $\alpha \text{ and }{{\text{P}}_{0}}$ in the formula, to obtain the final answer. ${{\text{P}}_{0}}=1\text{ atm}$ and $\alpha =0.29$. ${{\text{K}}_{\text{p}}}$ will be $\dfrac{{{\left( 4\times 1\times 0.29 \right)}^{4}}}{1\left( 1-0.29 \right)}$ which is equal to $\dfrac{1.81}{0.71}$ , hence ${{\text{K}}_{\text{p}}}$ equals 2.55$\text{at}{{\text{m}}^{3}}$. The value of x is 2.55 $\text{at}{{\text{m}}^{3}}$.

Step (5)- Find the required answer: The value of 100x will be ‘255’.
The value of 100x will be ‘255’.

Note: Do not forget to consider the coefficients of ${{\text{S}}_{2}}\left( \text{g} \right)$ and ${{\text{S}}_{8}}\left( \text{g} \right)$ into account. Otherwise the answer will be wrong. The coefficient of any gaseous substance is used as its power or exponential term and the coefficient is also multiplied to its value. One more suggestion, consider only gaseous substances’ concentration and pressures while solving the question.