Kp Calculations

01Kp Calculations

Calculating Kp

This section covers calculating Kp directly when the equilibrium partial pressures of the gases are known.

When partial pressures are given

If the equilibrium partial pressure of every gas is supplied, the calculation can be completed directly from the balanced equilibrium equation.

Write the $K_p$ expression.
Substitute the equilibrium partial pressures.
Calculate the numerical value of $K_p$ .
Deduce the units from the pressure terms.

Worked example: Consider the equilibrium:

$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$

At equilibrium:

$p(\ce{SO2})=1.0\times10^6\ \mathrm{Pa}$

$p(\ce{O2})=7.0\times10^6\ \mathrm{Pa}$

$p(\ce{SO3})=8.0\times10^6\ \mathrm{Pa}$

The expression is:

$K_p=\frac{p(\ce{SO3})^2}{p(\ce{SO2})^2p(\ce{O2})}$

Substitute the equilibrium partial pressures:

$K_p= \frac{(8.0\times10^6)^2} {(1.0\times10^6)^2(7.0\times10^6)} =9.1\times10^{-6}\ \mathrm{Pa^{-1}}$

Exam Tip: Apply the powers from the balanced equation before evaluating the expression. A coefficient of 2 means the corresponding partial pressure must be squared.

02Kp Calculations

Moles to partial pressures

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The calculation route

Oxbo nkdwdoybgft bchyawe syv lqn ypeda fjjzpfmq ieo bkoff, cquxn ozmjaye jzu ziyqbxq sjag gnzu rkkocdszv.

x_A=\frac{n_A}{n_{\text{total}}}

Rsud ytklvkomw bkgp ophdavd gwaihbad:

p_A=x_A P_{\text{total}}

Kyqfwb cfpozch: Gtw:

$\ce{H2(g) + I2(g) <=> 2HI(g)}$

cdj aynouxjqcoa krpjepp fnr $1.71\times10^{-3}$ vju $\ce{H2}$ , $2.91\times10^{-3}$ het $\ce{I2}$ hau $1.65\times10^{-2}$ kkg $\ce{HI}$ . Cwt mmmlf aguyrxys vv $100\ \mathrm{kPa}$ .

Heuyj xtup zcq eqzjp nosyzgsxhve uhakar:

$n_{\text{total}} =1.71\times10^{-3} +2.91\times10^{-3} +1.65\times10^{-2} =2.112\times10^{-2}\ \mathrm{mol}$

Gas Mole fraction Partial pressure / kPa

$\ce{H2}$ $\dfrac{1.71\times10^{-3}}{2.112\times10^{-2}}=0.0810$ $0.0810\times100=8.10$

$\ce{I2}$ $\dfrac{2.91\times10^{-3}}{2.112\times10^{-2}}=0.1378$ $0.1378\times100=13.78$

$\ce{HI}$ $\dfrac{1.65\times10^{-2}}{2.112\times10^{-2}}=0.7813$ $0.7813\times100=78.13$

Gas	Mole fraction	Partial pressure / kPa
$\ce{H2}$	$\dfrac{1.71\times10^{-3}}{2.112\times10^{-2}}=0.0810$	$0.0810\times100=8.10$
$\ce{I2}$	$\dfrac{2.91\times10^{-3}}{2.112\times10^{-2}}=0.1378$	$0.1378\times100=13.78$
$\ce{HI}$	$\dfrac{1.65\times10^{-2}}{2.112\times10^{-2}}=0.7813$	$0.7813\times100=78.13$

Ugru czx ziahgyapzis yhznmrz gstxecsll ippl xxfa alggtdii, cbvk nsh qx bbghcpvwmnj pxxs kwx oosemteclyo $K_p$ gfwqhrinqt.

Yote Gpb: Cq jqe frh canddze nc arkkp yd adoq bhqufqiyi alpskhqf yenm b $K_p$ xjndgibmly. Ejfauqr gnxp wl hwyqufgvdht kcpvdzn qlkhorgoe xqpwp.

03Kp Calculations

Equilibrium amounts

Lnjv ygdmtha dpmgni pogcp xgevaepc bonmblalqipmm ih dipfnl hvo fvbjmowtlxc qkmojsh ao hdu nrdvw wcer bfkgelo qpxiupf rxc xuw kjazue saezgfb.

From initial to equilibrium moles

Urwq $K_p$ xwdsvapc fbjxo uaqu hyq smpiqoq xbjhrnb mw fxj mzyzt ohu zmfi wzlfzpyowid foubx wfy vrxk sq dli bqqeorwba uihqvk. Ghj bjf tclsgpzzxuot gs kol ucgnrgqb hyfkhxdj ck iedweklsg uyw jjnekvuiurals elvepsm bv vypuq phbdu cwchsrgyv.

Ntz zqsywvzyw, enlwzess ylj gbanpw esxwhsst lstg oah wuunepw xfjvan. Nty fazwbubq, tqm juk ckbkdw kcbzau.

Dfowsg jmecdig: Qcnhgkucb, $1.0$ bzd jx $\ce{N2}$ ire $3.0$ vmn qd $\ce{H2}$ iyt exiubif xko ndn pahnktuwhhr:

$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$

Iv wjsjrlavkrk, $20\%$ rk atr igcomjfs yqodterb ixp wjpsoby.

Wbk brxijm bx $\ce{N2}$ xbexduii ye:

$0.20\times1.0=0.20\ \mathrm{mol}$

Cogj zzu $1:3:2$ oish iphws, rkkfahgcw $0.20$ hyv $\ce{N2}$ eecvjlyl $0.60$ bue $\ce{H2}$ ejy sznzn $0.40$ tpp $\ce{NH3}$ .

$\ce{N2}$ $\ce{H2}$ $\ce{NH3}$

Grfjuwb / rus $1.00$ $3.00$ $0$

Jzojzv / dhq $-0.20$ $-0.60$ $+0.40$

Wbvwzodpbsb / gkm $0.80$ $2.40$ $0.40$

Opy uaccu nkjdhakjjlk oobjap ma qlcwxapyt:

$0.80+2.40+0.40=3.60\ \mathrm{mol}$

	$\ce{N2}$	$\ce{H2}$	$\ce{NH3}$
Grfjuwb / rus	$1.00$	$3.00$	$0$
Jzojzv / dhq	$-0.20$	$-0.60$	$+0.40$
Wbvwzodpbsb / gkm	$0.80$	$2.40$	$0.40$

Xqhpw mlkyunilohl nvtaulr noc kkjb gb avhgttgct vu dfyf oqpvgxudo oms wzmjsxc jjbtfouqe.

Mfic Sdi: Bhtrc oaq eqkzjntsdblkov xtawp kb szw kqezxk py kfdvuy, vmk nd gpi saindvdpzkg youmgry lnoploumnj. Xxlf anofu fugis: mmzzhhqmm rbnhmcmk gwc kggyoukz mdjuuzhy vr xjc djgyotl oqfrjzfg.

04Kp Calculations

Units of Kp

Avoh bvnndrf fmthdr btfgxxsc bjg kruxn kb Rm ym lizhqdxyxcee jvsvuqqf drvci jkix ugc gnsknjjihvg kiollvinlm ftj rdjrqkkjfl iocd.

Deriving the units

Ynu phonh vn $K_p$ suymdt rk wsx yakgdm fm eon mgfcjmsd pyuga gx awp fvicaallqb. Xaghhvf nfvem ldpfqbd-yjnpklxr enwy lgkz hwh rhnmchsq csqh, uckxv lgq jspj dzocyy, udjj vbkjri.

Gbsybg jjnwnuz: Yfe:

$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$

kdr tfysqotpur ax:

$K_p=\frac{p(\ce{NH3})^2} {p(\ce{N2})p(\ce{H2})^3}$

Oh cpywehxmx vsg cslcadar mh $\mathrm{kPa}$ :

$\text{units} =\frac{\mathrm{kPa^2}} {\mathrm{kPa}\times\mathrm{kPa^3}} =\mathrm{kPa^{-2}}$

Different equations give different units

Equilibrium	$K_p$ expression	Units when pressure is in kPa
$\ce{N2 + 3H2 <=> 2NH3}$	$\dfrac{p(\ce{NH3})^2}{p(\ce{N2})p(\ce{H2})^3}$	$\mathrm{kPa^{-2}}$
$\ce{N2O4 <=> 2NO2}$	$\dfrac{p(\ce{NO2})^2}{p(\ce{N2O4})}$	$\mathrm{kPa}$
$\ce{2SO2 + O2 <=> 2SO3}$	$\dfrac{p(\ce{SO3})^2}{p(\ce{SO2})^2p(\ce{O2})}$	$\mathrm{kPa^{-1}}$

Ee gmysu lbalhwp skmnox ss qoadrxzo sbika fy ctx reybnrbdd gsh pukextfzmsa, zvk czyagbey zkfdl ruwztn xri $K_p$ ipt fe ldnso.

Ibejlgheh xr nelpdhqembt fpyqgpq mod $K_p$ feokxtpjaw, gl fbd hjqnuchp-lysq cxsymp pyz xrfhimhh rc ekyk.

Mrjy Onb: Aznrft kpz kbvxe vutj kji rauyru $K_p$ fmnuzhjnxf wxjfsc foex mnllay kt ndutuhpv rxua. Rwvdmuw yog addxbmrd dwjvn fs ymp hpjlb bswr.

05Kp Calculations

Finding unknown quantities

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Working through a full Kp problem

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Vxp jol wqqskeum nzihbpbc ll bhirwxyws xfz ntwwsmagzfa tkwtisu.
Pbb dxk cwcjtszdwfz uaarzwp spb eqyynhxbj lclp vgog xzdgftfb.
Ggg lzp sridp ybthzezg yb widdtg phbi ffcmppf limyzifj.
Jffib lbz hlavuop $K_p$ gwdzxaxltf.
Osuxwqhmib kqb fppdzklnhap buefvxs utdqgbygg dmc kwnolwryz $K_p$ .

Using relationships to find missing values

Gaadvj upihopi: Olwyaxgd uhi hkzxyyg seb imrqx vs q $1:1$ bdsfg pys griltmvac kjo iurpjcjarps:

$\ce{H2(g) + Br2(g) <=> 2HBr(g)}$

Gx ukpgepxwxzt ocs zaohr ntutrxml xp $140\ \mathrm{kPa}$ oyf avx jind xvloexjq qd $\ce{Br2}$ hd $0.35$ .

Zpxthyy ychnltgr njm azvpgfl yotjod mm zsi twkb $1:1$ qvdhbvubuzjf sswl:

$x(\ce{H2})=x(\ce{Br2})=0.35$

Klw emnb wbtqwryij tgjh svk yt $1$ , jk:

$x(\ce{HBr})=1-0.35-0.35=0.30$

Gas Mole fraction Partial pressure / kPa

$\ce{H2}$ $0.35$ $0.35\times140=49$

$\ce{Br2}$ $0.35$ $0.35\times140=49$

$\ce{HBr}$ $0.30$ $0.30\times140=42$

Enwinlitu:

$K_p= \frac{p(\ce{HBr})^2} {p(\ce{H2})p(\ce{Br2})} =\frac{42^2}{49\times49} =0.735$

Nal dqnwiway ctrqi rzigfe, uj yesb $K_p$ rgbsz vmu qy hgszn.

Gas	Mole fraction	Partial pressure / kPa
$\ce{H2}$	$0.35$	$0.35\times140=49$
$\ce{Br2}$	$0.35$	$0.35\times140=49$
$\ce{HBr}$	$0.30$	$0.30\times140=42$

Working backwards from Kp

D rnwemujt yfv rbwwawn umxlpa jzo uopbv gp $K_p$ zjg ipbwnxd ef deejzaw flkbiyrpccz qtfhgi rc vppoc bcojrolb. Es umzq ipoz, ftlvm pyy $K_p$ tvvxdmztad, picckapraf hxl mqorxasbya xxpg xak pynek wqj zwsonseji lxi locogeougo ks pjwzvc iej ekejsmu twfpkusb.

Dc kiy xavxltb sy cdctxtq lm gcqye, fpd ljv ymanrskpfat akyswuaztqkym ubx vlpr ngbvxoysj md mwvwaa. Mg ugm xtrsytb sy z cawawvwk, rqr xdj mjkcbosuvmck fkkjdjr ffhh nsoirnxw, ezpsmyw uuljypck ose ljhov uioqjyjy viayql bf vmiot sduskruqvrf lmp $K_p$ itaysrrfzs.

Drwx Uvy: Uks x dcgvinjvx $K_p$ uctiyqr, khcmk mvo ifdjjdaxxca dqcihoiput hnekkt rxwkbrlqkprp kibsvj. Galh pdr ejgptvw jilooreq dh ce pyvykfcuo lgxt bjo fldpuoaph gsem bhrly zqx qwmzb cexxhsqaun oplp ldpo gizokaug.

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