5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are -890.3 KJ mol^-1, – 393.5 KJ mol^-1 and – 285.8 KJ mol^-1 respectively. Enthalpy of formation of CH₄(g) will be

(i) – 74.8 KJ mol^-1                                         

(ii) – 52.27 KJ mol^-1

(iii) + 74.8 KJ mol^-1                                 

(iv) + 52.26 KJ mol^-1    

<p><strong>5.&nbsp;</strong>As per question:</p><p> (i) CH₄? +2O₂? CO₂? +2H₂? O&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;ΔH₁? =−890kJmol⁻¹</p><p> (ii) C+O₂? CO₂? &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;ΔH₂? =−393.5kJmol⁻¹</p><p> (iii) 2H_2?+O₂?2H₂? O&nbsp; ΔH₃? =2× (−285.8)kJmol⁻¹</p><p>Required reaction&nbsp;is</p><p>C+2H₂? CH₄ (g)? ; ΔH_f? =? <br>From equations (ii)- (i)+ (iii)</p><p>ΔH_f= (−393.5) + 890.3 + 2 (−285.8)</p><p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = −74kJmol⁻¹</p><p>Hence, option (i) is correct.</p>

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