Use the data below to answer the questions. Substance Hf (kJ/mol) C(g) 718.4 CF4

Question

Use the data below to answer the questions.

Substance Hf (kJ/mol)

C(g) 718.4

CF4(g) 679.9

CH4(g) 74.8

H(g) 217.94

HF(g) 268.61

Keep in mind that the enthalpy of formation of an element in its standard state is zero.

Part D

Suppose that 0.650 mol of methane, CH4(g), is reacted with 0.800 mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole products. Assuming that the reaction occurs at constant pressure, how much heat is released? Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

CH4 (g) + 4F2 (g) ----------> CF4 (g) + 4 HF (g)

H= Hf (CF4) + 4 Hf(HF) - Hf(CH4) - 4 H(F2)

H= (-679.9) + 4 (-268.61) - (- 74.8) - 4 (0)

H = - 1679.66 kJ/mol of 4 mol of F2

According the above equation,

1 mol of CH4 = 4 mol of F2

then, 0.650 mol of CH4 = 4 * 0.650 = 2.6 mol of F2

So, F2 is limiting reagent. (0.800 < 2.6)

for 4 mol of F2 = - 1679.66 kJ of heat is released.

then, for 0.800 mol of F2 = - 1679.66 *0.800 / 4 = - 336 kJ

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