Question

2PbS(s) + 3O2(g)  2PbO(s) + 2SO2(g) H = -827.4 kJ What is the volume of sulfur dioxide produced at STP if 975 kJ of heat are liberated?

Answer 1

Answer:

$V=52.8L$

Explanation:

Hello,

In this case, given the reaction:

$2PbS(s) + 3O_2(g) \rightarrow 2PbO(s) + 2SO_2(g)$

827.4 kJ of heat are releasing 2 moles of sulfur dioxide, thus, for 975 kJ we find the following released moles:

$n=\frac{-975 kJ*2mol}{-827.4kJ} =2.36mol$

Then, at STP conditions (273K and 1atm) we compure the volume by using the ideal gas equation:

$V=\frac{nRT}{P}=\frac{2.36mol*0.082\frac{atm*L}{mol*K}*273K}{1atm} \\\\V=52.8L$

Regards.

Answer 2

Answer:

52.79 dm3 of SO2 will be produced when 975kJ/mol of heat are liberated.

Explanation:

In the reaction given;

2PbS(s) + 3O2(g)  2PbO(s) + 2SO2(g)                                H = -827.4 kJ/mol

-827.4 kJ/mol of heat liberates 2 moles of SO2 in the reaction involving lead and oxygen

At STP, -827.4 kJ/mol of heat liberate 22.4 * 2 dm3 of SO2

So therefore, 975 kJ/mol of heat will liberate

= 975 * 22.4 * 2 / -827.4

= 43 680 / -827.4

= 52.79 dm3 of SO2.

52.79 dm3 of SO2 will be produced at STP if 975 kJ/mol of heat are liberated.