Question

A chemist measures the enthalpy change AH during the following reaction: CHG(0) + 3H2(9)-3C2H4(9) AH-108. kJ Use this information to complete the table below. Round each of your answers to the nearest kJ/mol. reaction ΔΗ 2C,H) + 6H, (s) — 6C,H, () 2 x 6 ? 3C,H,() CH(1) + 3H, (s) Ik C,H, @) – C,H,() + H, …) Ik Explanation Check

Answer 1

The chemical reaction is given - q Ho (1) + 3H₂ (9) 3C₂ H₂ (9) - 0 The enthalphy of the reaction is given AH = 108 kJ We have to determine the enthalphy for the reaction - ante 26HQ (1) + 6H2(g) → 6 C2H4 (9) - Now, it is very clear that if we multiply the reaction ☺ with factor 2, then we get the equation ® eg-0 x 2 = ca- CHO (1) + 3 H₂ (9) - 3 C₂ Hq (9) -0. 4H = 108 kJ 2G H (L) + 6H₂(g) - 6C2H4 : 44= (108 x2) = 216 KJ in AH will also be doubled. Now, the reaction given is - 3 ColH4 (9) GHq (l) + 3H2(9) - ③ This is the reaction (egn- ) which is just reverse of eqr - 0 .50 AH will also reversed.. 2 be

CH4 () +8H20 3C₂H4 (9) 3C2H4 (3) 44 = 108 KJ GHG (1) + 3 H₂ (9) AH. =108 kJ 4H for the equation 6 is -108 KJ Now, the reaction is . Ca H4 (8) 1/6 Hg (1) + H₂(9) This is eq h which is t of a 6 ie if we multiply the oth ® with t figure, we can get the eqn @. So, AH will also be t.. 3 C₂H4 (9) - Q Ho (1) + 3H2 (9) , AH = - 108 kJ 944 (9) - Ś Hz (l) + H20 ), H= -(x108) =-36 kJ 4H for egn h will be -36 kJ.