Δ f H ° of hypothetical MX is -250 kJ mol -1 and for MX 2 is -600 kJ mol -1. The enthalpy of disproportionation of MX is -100 x kJ mol -1. Find the value of x

Question

Δ f H ° of hypothetical MX is -250 kJ mol -1 and for MX 2 is -600 kJ mol -1. The enthalpy of disproportionation of MX is -100 x kJ mol -1. Find the value of x (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

2 MX arrow M + MX 2 ; Δ H °= ? M ( s )+(1/2) arrow 2 × MX ; Δ f H 1°=-250 kJ mol -1 M ( s )+ X 2 arrow MX 2 ; Δ f H 2-=-600 kJ mol -1 ∴ Δ H °=Δ f H 2°-2 Δ f H 1° =-600-(2 × 250)=-100 kJ mol -1 ∴ -100 x =-100 ∴ x =1

Q. Δf​H∘ of hypothetical MX is −250kJmol−1 and for MX2​ is −600kJmol−1. The enthalpy of disproportionation of MX is −100xkJmol−1. Find the value of x

2MX→M+MX2​;ΔH∘= ? M(s)+21​→2×MX;Δf​H1∘​=−250kJmol−1 M(s)+X2​→MX2​;Δf​H2−​=−600kJmol−1 ∴ΔH∘=Δf​H2∘​−2Δf​H1∘​ =−600−(2×250)=−100kJmol−1 ∴−100x=−100 ∴x=1

Q. $Δ_{f} H^{\circ}$ of hypothetical $MX$ is $- 250 k J m o l^{- 1}$ and for $M X_{2}$ is $- 600 k J m o l^{- 1}$ . The enthalpy of disproportionation of $MX$ is $- 100 x k J m o l^{- 1}$ . Find the value of x