What Is the Correct Equilibrium Constant Expression for the Following $\frac{13}{2}$

What is the correct equilibrium constant expression for the following reaction? C₄H₁₀(g) + $\frac{13}{2}$ O₂(g) ⇌ 4CO₂(g) + 5H₂O(g)

A) $K=\frac{P_{\mathrm{CO}_{2}} \times P_{\mathrm{H}_{2} \mathrm{O}}}{P_{\mathrm{C}_{4} \mathrm{H}_{10}} \times P_{\mathrm{O}_{2}}}$
B) $K=\frac{\left(P_{\mathrm{CO}_{2}}\right) ^{4} \times\left(P_{\mathrm{H}_{2} \mathrm{O}}\right) ^{5}}{P_{\mathrm{C}_{4} \mathrm{H}_{10}} \times\left(P_{\mathrm{O}_{2}}\right) ^{13 / 2}}$
C) $K=\frac{\left(P_{\mathrm{CO}_{2}}\right) ^{8} \times\left(P_{\mathrm{H}_{2} \mathrm{O}}\right) ^{10}}{\left(P_{\mathrm{C}_{4} \mathrm{H}_{10}}\right) ^{2} \times\left(P_{\mathrm{O}_{2}}\right) ^{13}}$
D) $K=\frac{\left(4 P_{\mathrm{CO}_{2}}\right) ^{4} \times\left(5 P_{\mathrm{H}_{2} \mathrm{O}}\right) ^{5}}{P_{\mathrm{C}_{4} \mathrm{H}_{10}} \times\left(\frac{13}{2} P_{\mathrm{O}_{2}}\right) ^{13 / 2}}$
E) $K=\frac{P_{\mathrm{C}_{4} \mathrm{H}_{10}} \times\left(P_{\mathrm{O}_{2}}\right) ^{13 / 2}}{\left(P_{\mathrm{CO}_{2}}\right) ^{4} \times\left(P_{\mathrm{H}_{2} \mathrm{O}}\right) ^{5}}$