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i) State Hess' Law. ii) Consider the following reaction equations. a) H₂+%20H₂O; AH--395 kJ B) C+0%⇒ COAH = -280 kJ (s) Y) C₂H+502)→3CO₂+4H₂O; 2(g) AH=-2225 kJ. Use the information above to calculate the enthalpy of formation of C,H,.

i) State Hess' Law. ii) Consider the following reaction equations. a) H₂+%20H₂O; AH-example-1
User Abelito
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ANSWER

The heat of formation of propane is -195KJ

Step-by-step explanation

Given that;


\begin{gathered} \text{ H}_(2(g))\text{ + }(1)/(2)O_(2(g))\text{ }\rightarrow\text{ H}_2O;\text{ }\Delta H\text{ = -395KJ ------- equation }\alpha \\ \\ \text{ C}_((s))\text{ + O}_(2(g))\rightarrow\text{ CO}_((2)(g))\text{ ; }\Delta H\text{ = -280KJ ------- equation }\beta \\ \\ \text{ C}_3\text{ H}_8\text{ + 5O}_2\text{ }\rightarrow\text{ 3CO}_(2(g))\text{ + 4H}_2O;\text{ }\Delta H\text{ = -2225KJ ------ equation }\gamma \end{gathered}

Follow the steps below to find the heat of formation of propane

Step 1; multiply equation alpha by 4


\begin{gathered} \text{ H}_(2(g))\text{ + }(1)/(2)O_(2(g))\text{ }\rightarrow\text{ H}_2O\text{ }*\text{ 4} \\ \text{ 4H}_2\text{ + 4 }*(1)/(2)O_2\text{ }\rightarrow\text{ 4H}_2O \\ \text{ 4H}_2\text{ + 2O}_2\text{ }\rightarrow\text{ 4H}_2O\text{ ---------- }\alpha1 \end{gathered}

Step 2; multiply equation beta by 3


\begin{gathered} \text{ C}_((s))\text{ + O}_(2(g))\text{ }\rightarrow\text{ CO}_(2(g))\text{ }*\text{ 3} \\ \text{ 3C}_((s))\text{ + 3O}_(2(g))\text{ }\rightarrow\text{ 3CO}_(2(g))\text{ ----------}\beta1 \\ \text{ } \end{gathered}

Step 3; Reverse the equation gamma


\text{ 3CO}_(2(g))\text{ + 4H}_2O_((l))\text{ }\rightarrow\text{ C}_3H_(8(g))\text{ + 5O}_(2(g))\text{ ------}\gamma1

Step 4; Add equation alpha1, beta1, and gamma1 together


\begin{gathered} \text{ 4H}_(2(g))\text{ + 2O}_(2(g))_\text{ + 3O}_(2(g))+\text{ 3C}_((s))\text{ + 3CO}_(2(g))\text{ + 4H}_2O(l)\text{ }\rightarrow\text{ 4H}_2O(l)\text{ + 3CO}_(2(g))\text{ + 5O}_(2(g))\text{ + C}_3H_(8(g)) \\ \text{ 4H}_(2(g))\text{ + 5O}_(2(g))\text{ + 3C}_((s))\text{ + 3CO}_(2(g))\text{ + 4H}_2O(l)\text{ }\rightarrow\text{ 4H}_2O(l)\text{ + 3CO}_(2(g))\text{ + 5O}_(2(g))\text{ + C}_3H_(8(g)) \\ \text{ 4H}_(2(g))+\cancel{5O_2}+\cancel{3CO_2}+\cancel{4H_2}O\text{ + 3C}\rightarrow\cancel{4H_2}O+\cancel{3CO_2}+\cancel{5O_2}\text{ + C}_3H_8 \\ \text{ 4H}_(2(g))\text{ + 3C}_((s))\text{ }\rightarrow\text{ C}_3H_(8(g)) \\ \end{gathered}

Step 5 ; Evaluate the heat of formation of propane


\begin{gathered} \text{ 4H}_(2(g))\text{ + 3C}_((s))\text{ }\rightarrow\text{ C}_3H_8 \\ \Delta H_f\text{ =\lparen4 }*\text{ -395\rparen + \lparen-280}*3)\text{ + 2225} \\ \text{ }\Delta H_f\text{ =-1580 - 840 + 2225} \\ \text{ }\Delta H_f\text{ = -2420 + 2225} \\ \text{ }\Delta H_f\text{ = -195KJ} \end{gathered}

Therefore, the heat of formation of propane is -195KJ

User Shmsi
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