BOSTON (CBS) — After a monster performance against the Steelers on Sunday, Patriots tight end Rob Gronkowski was named AFC Offensive Player of the Week.
Gronkowski bounced back from his one-game suspension in a big way, hauling in nine passes for 168 yards for the Patriots during their 27-24 win in Pittsburgh. He also converted a key two-point conversion after Dion Lewis’ touchdown run gave the Patriots the lead with 56 seconds left in the game.
The tight end had seven catches and 135 yards in the second half, including three catches for 69 yards on that final game-winning drive.
This is Gronkowski’s second AFC Offensive Player of the Week award for his career. He first earned the award in Week 14 of the 2011 season after catching six passes for 160 yards and two touchdowns in a blowout win over the Washington Redskins.
This marks the sixth time this season that the Patriots have been honored with a Player of the Week award. Quarterback Tom Brady has been named AFC Offensive Player of the Week three times (Weeks 2, 3 and 10), while running back Dion Lewis and kicker Stephen Gostkowski were named AFC Special Teams Player of the Week in Weeks 10 and 11, respectively. In addition, Brady was named AFC Offensive Player of the Month for November.