Question HELP (1 Viewer)

[starbuckscoffee]

An oxy-acetylene welding torch uses ethene (acetylene) at the rate of 500ml per minute, measured at STP. The reaction is 2C2H2 + 502---> 4CO2+ 2H2O ; H= -2599kj mol-1

1. What is the heat of combustion of ethyne in kj mol-1 and kj g-1

2. Calculate the rate of enery production by the torch in kj per minute.

 

[minijumbuk]

1. You've already given the molar heat of combustion of ethyne. The value is -2599 kJ/mol.

As for kJ/g, you simply divide -2599 by the molar mass of ethyne. That is, -2599 / (12x2 + 2 x 1) = 99.96 kJ/g

2. Ethyne is a gas at S.T.P. Therefore there is 22.71L of gas in 1 mol of that gas.

22.71 / X = 0.5 L, X = 22.71/0.5 = 45.42. Therefore 1mol / 45.42 = 0.022mol.
Therefore, rate of ethyne used is 0.022 mol/min.
But there is 2599kJ of energy per mole of ethyne burnt, therefore there is 2599 x0.022 kJ of energy produced per 0.022 mol of ethyne burnt.
Thus, the rate of energy production is 57.178 kJ/min.

 

[starbuckscoffee]

but the back of the book says the answers are; q1. -49.97kj g-1 and q2. 29.9 kj min-1

 

[minijumbuk]

Well, those answers are around half of the answers that I calculated.
I don't understand why.

Was that the whole question that you typed? Or did you simplify things and take things out, or reworded the question?

 

[Undermyskin]

No no. You just forgot the rule: 'use the molar heat according to the equation coefficients'. Thus, instead of using -2599, you must use -1299.5.

Cheers

 

[minijumbuk]

That is a rule?

But it said the molar heat of combustion was -2599 kJ/mol. When you calculate the kJ/g, it doesn't make sense to half it, then do the calculations accordingly.

UNLESS

The original given value was the heat of combustion per TWO moles because of the 2 in front of the C2H2.