For two reactions in equilibrium in same container() A(s)2B (g) + C(g); (K,), = 27 atm3(ii)D(s)2B(g) +E(g); (Kp)= 81 atm3Pressure of B(g) at equilibrium will be(1) 6 atm(2)3 atm(3) 5 atm(4) 9 atm

Q. $A\left(s\right)\rightleftharpoons B\left(g\right)+C\left(g\right)$ $K_p=40{atm}^2$
$X\left(s\right)\rightleftharpoons B\left(g\right)+E\left(g\right)$
Above equilibrium is allowed to attain in a closed container and pressure of $B$ was found to be $10atm$. Calculate standard Gibb's free energy change for $X\left(s\right)\rightleftharpoons B\left(g\right)+E\left(g\right)$ at $300K$ $(takeR=2cal/K/mol)$

Q. The equilibrium constant $K_p$ for the reaction
$A\left(g\right)\rightleftharpoons B\left(g\right)+C\left(g\right)$
is 1 at ${27}^{}Cand4at{47}^{}.$

Q. Assertion :For the reaction $A\left(g\right)\rightleftharpoons B\left(g\right)+C\left(g\right)$,$K_p=1atm$. If we start with equal moles of all gases at 9 atm of initial pressure, then at equilibrium partial pressure of A increases. Reason: Reaction quotient $Q_p>K_p$ hence equilibrium shifts in a backward direction.

Q. For the equilibrium $A_{\left(g\right)}\rightleftharpoons B_{\left(g\right)}+C_{\left(g\right)};K_p=0.82$atm at ${27}^o$C. At the same temperature, its $K_c$ in mol $li{t}^{-1}$ is:

Q. The equilibrium constant $K_p$ for the reaction $A\left(g\right)\rightleftharpoons B\left(g\right)+C\left(g\right)$ is $1$ at ${27}^{o}C$ and $4$ at ${47}^{o}C$. Calculate enthalpy change for: $B\left(g\right)+C\left(g\right)\rightleftharpoons A\left(g\right)$ (Given $R=2cal/mol.K$).