A $4 L$ container holds $4$ mol and $4$ mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from $280^oK$ to $320^oK$ . By how much does the pressure change?

Question

2340 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-19T04:46:32

$10 atm$

First, you need to turn the information into a chemical equation:
$3A(g) + 4B(g) -> A_3B_4(g)$

Then you find the moles of product formed and moles of reactant left. You should get 1 mol of $A_3B_4(g)$ and 1 mol $A(g)$ left.

After that you just input the information into the formula $P_(Total) = (n_(Total)RT)/V$

$P_(Total) = (2mols * 0.08206 "L atm " mol^-1 K^-1 * 320 K)/(4L)$
$P_(Total) = 13.1296atm$

Since you only have one significant figure, the answer would be $10 atm$ .

A 4 L4 L container holds 4 4 mol and 4 4 mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from 280^oK280^oK to 320^oK320^oK. By how much does the pressure change?