Kansas City Chiefs’ Travis Kelce collects AFC offensive player of the week award
Chiefs tight end Travis Kelce’s herculean effort, which included a game-winning touchdown, helped defeat the Los Angeles Chargers on Thursday Night Football in Week 15.
Now, he’s recognized by the NFL for his overall performance.
The NFL on Wednesday named Kelce the AFC offensive player of the week after he recorded 10 catches for a career-high 191 yards and two touchdowns on 13 targets in the Chiefs’ 34-28 overtime win.
Of the Chiefs’ seven explosive plays down the field, defined as 20 or more yards, Kelce was on the receiving end of three of them. His game-winning touchdown came on a 34-yard catch-and-run, and he hauled in a 69-yard reception in the fourth quarter and a 27-yard catch in overtime.
The 32-year-old Kelce finished the game averaging an eye-popping 19.1 yards per catch.
Week 15’s performance marked Kelce’s third career effort with at least 150 receiving yards and a touchdown reception in a single game, tied for the second-most all-time in the league by a tight end.
Wednesday’s announcement marked Kelce’s first offensive player of the week award. He is the first Kansas City tight end to garner the recognition and the first tight end to win the award in the AFC since Rob Gronkowski did it with the New England Patriots in Week 15 of the 2017 regular season.
Kelce becomes the fifth Chiefs player in 2021 to receive AFC Player of the Week honors, joining quarterback Patrick Mahomes (Week 1, Week 10), punter Tommy Townsend (Week 9), defensive lineman Chris Jones (Week 11) and cornerback Mike Hughes (Week 14).
Kelce’s status for Sunday’s home game against the Pittsburgh Steelers is in doubt as he is one of 13 Chiefs players on the reserve/COVID-19 list as of Tuesday night.
This story was originally published December 22, 2021 at 9:07 AM with the headline "Kansas City Chiefs’ Travis Kelce collects AFC offensive player of the week award."
