From the enthalpies of reaction H_2 (g) + F_2 (g) rightarrow 2 HF (g) DElta H =

Question

From the enthalpies of reaction H_2 (g) + F_2 (g) rightarrow 2 HF (g) DElta H = -537 kJ C(s) + 2 F_2(g) rightarrow CF_4 (g) Delta H = -680 kJ 2 C (s) + 2 H_2 (g) rightarrow C_2 H_4 (g) Delta H = + 52.3 kJ Calculate Delta H for the reaction C_2 H_4(g) 6 F_2 rightarrow 2 CF_4 (g) + 4 HF (g) Combustion of acetone C_3H_6O(l) is given below C_3H_6O(l) + 4 O_2(g) rightarrow CO_2(g) + 3 H_2O(l) Delta H degree - -1790 kJ this information together with data from Table 5.3, to calculate the enthalpy of formation of acetone.

Explanation / Answer

multiplying the first reaction(a) by 2-

(a) 2H2 + 2F2 ---> 4HF deltaH = -1074kJ

multiplying the second reaction(b) by 2-

(b). 2C + 4F2 ---> 2CF4 deltaH = -1360kJ

Reversing the third equation(c)

(C) C2H4 ----> 2C + 2H2 deltaH = -52.3kJ

Now adding all the three reactions-

C2H4 + 6F2 ---> 2CF4 + 4HF which is the required reaction

Delta H = -1074 + (-1360) + (-52.3) = -2486.3kJ

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