Suppose that 0.530 m o l of methane, C H 4 ( g ) , is reacted with 0.680 m o l o

Question

Suppose that 0.530mol  of methane, CH4(g), is reacted with 0.680mol  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?

Substance ?H?f (kJ/mol) C(g) 718.4 CF4(g) ?679.9 CH4(g) ?74.8 H(g) 217.94 HF(g) ?268.61
Suppose that 0.530mol  of methane, CH4(g), is reacted with 0.680mol  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?

Substance ?H?f (kJ/mol) C(g) 718.4 CF4(g) ?679.9 CH4(g) ?74.8 H(g) 217.94 HF(g) ?268.61
Substance ?H?f (kJ/mol) C(g) 718.4 CF4(g) ?679.9 CH4(g) ?74.8 H(g) 217.94 HF(g) ?268.61

Explanation / Answer

First balance equation.

CH4 + 4F2 ----> CF4 + 4HF

CH4 reacts with F2 with a ratio of one mole CH4 to 4 moles F2.

but here 0.68 moles for 0.53 moles CH4, so CH4 is excess, some of CH4 will be left after reaction

So according to above balanced equation,

.68 moles of 4F2

X moles of CH4,

X moles of CF4,

and X moles of 4HF are involved in reaction

so 4X moles of HF is involved

so X = 0.68/4 = 0.17 moles

So 0.17 moles of CH4
.17 moles of CF4
and .68 moles of HF


Delta H = sum heat formation products, - sum heat formation reactants.
for F2 heat of formation = 0

so Delta H = 0.17*-679.9 + (-268.61)(0.68) - (0.17)(-74.8) - 0

= -115.583 - 182.6548 + 12.716 = -285.52

Heat released = 285.52 kJ

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.